One can do Euclidean geometry as a completely formal game of symbolic logic. Euclid's axioms are almost sufficient for this, except that they lack formal support for some "obvious" continuity properties such as
Given a circle with center $C$, and a point $A$ such that $|CA|$ is less than the radius of the circle. Then any straight line through $A$ intersects the circle.
David Hilbert, one of the pioneers of the formalist viewpoint, developed an axiomatic system that closes these gaps and allows any Euclidean theorem about a finite number of lines, circles, and points to be proved completely formally. One can work in this system without any reference to arithmetic or set theory, considering Hilbert's geometry axioms as an alternative to, say, ZFC as one's formal basis for one's reasoning.
(Edit: Oops, my history was slightly wrong. Hilbert's axiomatic system was not completely formal. Alfred Tarski later developed a completely formal system; what I say here about Hilbert rightfully ought to read Tarski instead.)
Granted, when one does that one doesn't necessarily think of lines and points as "really existing" in some Platonic sense -- after all, the basic idea of formalism is that no mathematical objects "really exist" and it's all just symbols on the blackboard that we play a parlor game of formal proofs with. But that does not mean that it is necessary, or even desirable, to consider points to be coordinate tuples while playing the game. Indeed, many mathematicians would probably agree that the points and lines of Hilbert's geometric axioms exist "in and of themselves" to at least the same extent that the sets of ZFC (or, for example, the real numbers) exist "in and of themselves".
There are some additional points about this state of things that have the potential to cause confusion, but do not really change the basic facts:
When I speak of Hilbert's axioms as an "alternative" to ZFC, I don't mean that they can be used as a foundation for all of mathematics they way ZFC is -- because they have not been designed to fill that role. I mean merely that they occupy the same ontological position in terms of which concepts one needs to already be familiar with in order to work with them. Perhaps "parallel" might be a better word than "alternative".
The rules of what consists valid formal proofs (in ZFC or geometry) are ultimately defined informally. One may construct a formal model of the rules, but that just punts the informality to the next metalevel, because reasoning about the formal model itself needs some sort of foundation.
When one does formalize the rules of formal proofs, one often does that in a set-theoretic setting. However, this doesn't mean that a different theory such as Hilbert's geometry "depends on" set theory in a fundamental way. Remember that the set theory we use to formalize logic can itself be considered a formal theory, and at some point we have to stop and be satisfied with an informal notion of proof (it cannot be turtles all the way down). And there is no good reason why there has to be a set-theoretic metalayer below the geometry before we reach the inevitable point of no further formalization.
This does not mean that formalizing the rules of formal geometric proof is a pointless exercise. Doing so can tell us things about the axiomatic system that cannot be proved within the formal system itself. In particular, if we formalize the axiomatic system within set theory, we get access to the very strong machinery of model theory to prove facts about the axiomatic system. This is where numbers and coordinates enters the picture, as described in Qiaochu's answer. Using these techniques, one can prove [as a theorem of set theory] that every formal statement about a finite number of lines, points and circles can be either proved or disproved using Hilbert's axioms.
Hilbert's system does not allow one to speak about indeterminate numbers of points in a single statements -- so one cannot prove general theorems about, say, arbitrary polygons. This is by design. The kind of reasoning usually accepted for informal proofs about arbitrary polygons is so varied that formalizing it would probably just end up being a more cumbersome way to do set theory, and there's not much fun in that.
Reading your situation, I can offer some advice from someone who is 33 years old and currently self-studying Apostol. I cannot speak to Spivak directly, but I imagine the things I have to say about Apostol will apply more or less equally to Spivak.
Apostol is a wonderful book, but would not be my choice as an introduction to either calculus or proof-based mathematics. My background differs somewhat from yours in that I took a full year of Calculus in high school, and have an engineering degree so took classes on ODE, linear algebra, and statistics (amongst some other courses that made significant use of mathematical techniques). That was years ago, so I've forgotten many things, but with that background I find Apostol challenging at times, and it certainly requires a lot of motivation to keep moving forward. Without someone to guide you, there are lots of places in a rigorous treatment of Calculus to get hopelessly confused. If you look at my profile you can see the questions (an expanding list) that I have asked from problems in Apostol to get a flavor for the sorts of exercises that the book contains. I think it would be very difficult to undertake the book without a good background in proving mathematical statements rigorously, and a strong notion of the ideas of Calculus. The theorems and definitions would just be too obtuse if you didn't already understand what was going. Apostol is not the place to learn these things.
That said, I would echo one of the comments that there is math to learn other than Calculus. Not to say Calculus shouldn't be on your agenda, just that there are beautiful and accessible areas of math that might appeal to you and are better for developing mathematical technique. I strongly recommend looking into books on elementary number theory (preferably something with complete solutions) or discrete math. These areas are completely accessible to you right now, provide a forum in which to apply some of the basic mathematical techniques you are learning, will really help you develop your skills with manipulating equations and "thinking mathematically," and will start you on the road to understanding the difference between solving equations at the algebra level and writing proofs.
As for actual book recommendations, a very straightforward number theory book with solutions is Jones & Jones. It's not a great book at all, but it has full solutions and is targeted at undergraduates. A very ambitious book might be Concrete Mathematics, much of which will be too advanced at the moment, but is a book that can grow with you. It contains wonderful discussions of very concrete (as the title might suggest) problems, and methods for working with sums and equations which are invaluable as you gain maturity.
Hope that helps.
Added: Also since you seem to be interested in mathematics generally, I would highly recommend taking the time to read some math history and "popular" math books. Math has a wonderfully rich and entertaining history. These can give great insight into problems that have been pivotal to the development of mathematics, and give glimpses to the sorts of things mathematicians are interested in. Also, many "pop" math books are about theorems or concepts that have fascinated mathematically minded people for centuries, so they are likely to fascinate you. Finally, they can be read right now while you are still developing your basic math skills, and can provide lots of motivation and inspiration.
Specific books I'd recommend are E.T. Bell's "Men of Mathematics" for math history, some of recent books on the Riemann hypothesis are good reads about an open problem (du Sautoy, The Music of the Primes and Derbyshire's "Prime Obsession" were both good), and I found "Symmetry and the Monster" by Ronan really inspiring (if you want to get a glimpse of what "modern" algebra is about).
Best Answer
Aleph numbers are used to denote infinite cardinalities.
Before we go any further, we need to discuss some basic facts about cardinalities.
Given sets $A$, $B$, we say $A \approx B$ exactly when there is a bijection $A \to B$. For every set $A$, there is a cardinal $|A|$. Every cardinal can be expressed as $|A|$ for some $A$. And for any sets $A, B$, we have $|A| = |B|$ if and only if $A \approx B$.
Given two cardinals $|A|$ and $|B|$, we say $|A| \leq |B|$ if and only if there exists some injection $A \to B$. We can verify that $\leq$ is transitive and reflexive using the fact that the identity map $1_A : A \to A$ is injective and using the fact that the composition of injective maps is injective. The Schroder-Bernstein theorem further states that $\leq$ is a partial ordering on cardinals. In other words, if $|A| \leq |B|$ and $|B| \leq |A|$ then $|A| = |B|$.
When $\kappa$ and $\lambda$ are cardinals, we say that $\kappa < \lambda$ if and only if both $\kappa \leq \lambda$ and $\kappa \neq \lambda$.
Using the axiom of choice, we can further prove that $\leq$ is a total ordering. That is, given any cardinals $\lambda$ and $\kappa$, either $\lambda \leq \kappa$ or $\kappa \leq \lambda$.
In fact, the axiom of choice allows us to prove that $\leq$ is a well-ordering. This means that given any nonempty collection of cardinals $F$, there is a smallest element of $F$.
There are plenty of obvious cardinals that we know how to deal with. These obvious cardinalities are the cardinalities of finite sets. It's easy for us to check which of two finite sets is larger, and many operations of "cardinal arithmetic" turn out to be ordinary math when we are working with finite cardinals.
But what about the infinite cardinals? Are there any cardinals which are not finite?
This is where we require the axiom of infinity, which allows us to prove that there is set $\mathbb{N}$ of all natural numbers. $\mathbb{N}$ is infinite, so $|\mathbb{N}|$ is larger than any finite cardinal.
We can also prove, using the axiom of countable choice, that if $B$ is any set which is not finite, then $|\mathbb{N}| \leq |B|$. This means that $|\mathbb{N}|$ is the smallest infinite cardinal.
Since $|\mathbb{N}|$ is an infinite cardinal and there are $0$ infinite cardinals smaller than $|\mathbb{N}|$, we define $\aleph_0 = |\mathbb{N}|$.
The next question is: are there any other infinite cardinalities? In order to answer this question, we require the axiom of power sets, which state that power sets exist.
We can use Cantor's Diagonalisation Argument to prove that if $S$ is any set, then $|S| < |P(S)|$. This proves in particular that there is no largest cardinality.
In particular, we can expand Cantor's argument to prove that given any set $S$ of cardinals, there is a cardinal $\kappa$ which is bigger than all the cardinals in $S$. This argument is somewhat technical.
Given any cardinal $\lambda$, we can consider the class $\mathcal{F}$ of all cardinals $\kappa$ such that $\kappa > \lambda$. We know that there is at least one $\kappa \in \mathcal{F}$, since there is no biggest cardinal. Therefore, because cardinals are well-ordered, there is a smallest element of $\mathcal{F}$. Call this element $\lambda^+$. Then $\lambda^+$ is the smallest cardinal $> \lambda$.
Now in particular, let's define $\aleph_1 = \aleph_0^+$. The only infinite cardinal $\kappa$ such that $\kappa < \aleph_1$ is $\kappa = \aleph_0$, since if we had some other $\kappa$, then we would have $\aleph_0 < \kappa < \aleph_1$ which contradicts the very definition of $\aleph_1 = \aleph_0^+$. That is, there is only 1 infinite cardinal smaller than $\aleph_1$.
Similarly, we can define $\aleph_2 = \aleph_1^+$, $\aleph_3 = \aleph_2^+$, and, in general, $\aleph_{n + 1} = \aleph_n^+$ for all $n \in \mathbb{N}$. We can show that for all $n$, there are exactly $n$ infinite cardinals smaller than $\aleph_n$.
So we've defined an infinite collection of infinite cardinals, $\aleph_n$ for all natural numbers $n$. The next question is: is there a cardinal which is even larger than all the $\aleph_n$s we've defined so far?
It turns out that we need the axiom of replacement to prove that the answer is "yes". Using the axiom of replacement, we can form the infinite set $S = \{\aleph_n \mid n \in \mathbb{N}\}$. But recall that given any set of cardinals $S$, we can form a cardinal $\kappa$ which is bigger than any element of $S$. So there is some cardinal which is even bigger than all the $\aleph_n$s where $n \in \mathbb{N}$. Call the smallest such cardinal $\aleph_\omega$. Note that there are exactly $|\omega| = |\mathbb{N}|$ cardinals smaller than $\aleph_\omega$.
Once we have $\aleph_\omega$, it's only natural to define $\aleph_{\omega + 1} = \aleph_\omega^+$. Note that there are exactly $|\omega + 1| = |\omega|$ infinite cardinals smaller than $\aleph_\omega$.
So once we get to infinite $\alpha$s, we can't just define $\aleph_\alpha$ as the infinite cardinal such that there are exactly $|\alpha|$ smaller infinite cardinals, since there are exactly $|\omega|$ infinite cardinals smaller than $\aleph_\omega$, and there are also $|\omega|$ infinite cardinals smaller than $\aleph_{\omega + 1}$.
Instead, we consider the fact that given an infinite cardinal $\kappa$, the set $S_\kappa := \{\lambda \mid \lambda $ an infinite cardinal, $\lambda < \kappa\}$ is well-ordered by $<$. What matters is not $|S_\kappa|$ - what actually matters is the order-type of $(S_\kappa, <)$.
To illustrate this, consider that $S_{\aleph_\omega} = \{\aleph_n \mid n \in \mathbb{N}\}$ has a very simple order $<$. We have $\alpha_a < \aleph_b$ if and only if $a < b$ as natural numbers. In other words, the order type of $(S_{\aleph_\omega}, <)$ is exactly the order type of the natural numbers, which is written as $\omega$.
By contrast, consider that $S_{\omega + 1} = S_\omega \cup \{\aleph_\omega\}$. This set's order type is $\omega + 1$, since the set can be broken down into $S_\omega$, which has order type $\omega$, and an element $\aleph_\omega$ which is bigger than all the elements of $S_\omega$.
Given every well-ordered set $S$, there is another well-order $type(S)$ such that $type(S)$ and $S$ are equivalent well-orders. Furthermore, $type(A) = type(B)$ if and only if $A$ and $B$ are well-orders. We say that the "ordinals" are the well-orders of the form $type(S)$.
So in general, we have the following definition:
Note that we can prove that if there are two cardinals $\kappa, \lambda$ such that $(S_\kappa, <)$ has the same order type as $(S_\lambda, <)$, then we can prove by induction on that order type that $\kappa = \lambda$. Thus, we know that there is at most one cardinal satisfying the definition of $\aleph_\alpha$ for all ordinals $\alpha$.
From this definition, it follows immediately that every infinite cardinal is an aleph. In particular, given an infinite cardinal $\kappa$, let $\alpha$ be $type(S_\kappa, <)$. Then it's immediate that $\kappa = \aleph_\alpha$.
We can prove using the axiom of replacement that for every ordinal $\alpha$, $\aleph_\alpha$ exists. The proof uses a technique called "ordinal induction".
We have thus discovered that the ordinals and infinite cardinals can be put into correspondence using the $\aleph$ function. This is a profound and philosophically satisfying result.