It is consistent without the axiom of choice that there are uncountable sets which cannot be well-ordered, and every infinite set whose cardinality is smaller is necessarily countable. It could also be the case that there are smaller cardinalities, but none which are strictly between $\aleph_0$ and the uncountable set.
Indeed there are three different definitions of successor cardinals when the axiom of choice fails (although the existence of one [for every set] implies choice; there is another which is provable without choice; and a third which is independent).
It is consistent, if so, that there are several successors to $\aleph_0$. It is always true that $\aleph_1$ is the successor of $\aleph_0$, and that it is the minimal aleph above it.
In particular it is consistent that the real numbers form such set.
See, for example, Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Edit: I have some free time so here are different variants of "successor". This is taken from Jech The Axiom of Choice (p. 163), the original definition is due to Tarski.
Let $\frak p$ and $\frak q$ be cardinals such that $\frak p<q$.
- $\frak q$ is the $1$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p\leq m\leq q$ then $\frak p=m$ or $\frak q=m$.
- $\frak q$ is the $2$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak p < m$ then $\frak q\leq m$.
- $\frak q$ is the $3$-successor of $\frak p$ if whenever $\frak m$ is such that $\frak m < q$ then $\frak m\leq p$.
Now we see that it is always true that $\aleph_1$ is a $1$- and $3$-successor of $\aleph_0$. However it is consistent that so are the real numbers themselves.
The assertion that $\aleph_1$ is a $2$-successor of $\aleph_0$ is equivalent to every uncountable set $X$ has an injection from $\omega_1$ into $X$. In fact just requiring that $\aleph_0$ has a $2$-successor is enough.
It is consistent that there are several $1$-successors to $\aleph_0$, and that there is a $1$-successor which is not $3$-successor.
For the real numbers it holds that if their cardinal is a $1$-successor of $\aleph_0$ then it is a $3$-successor of $\aleph_0$.
Interestingly, it is consistent that there is a proper class of $1$-successors to $\aleph_0$.
Best Answer
To prove the existence of $\aleph_1$ we use the concept of Hartogs number. The question asks, really, why are there uncountable ordinals, since $\aleph_1$ is by definition the least ordinal which is not countable.
Take a set of cardinality $\aleph_0$, say $\omega$. Now consider all the orders on $\omega$ which are well-orders, and consider the order isomorphism as an equivalence relation. The collection of all equivalence classes is a set.
Map every equivalence class to the unique ordinal which is order isomorphic to the members of the class. We now have a set and all its members are ordinals which correspond to possible well-ordering of $\omega$.
Let $\alpha$ be the union of the set defined above. We have that $\alpha$ is an ordinal, and that every ordinal below $\alpha$ is a possible well-ordering of $\omega$ (and therefore countable).
Suppose $\alpha$ was countable too, then $\alpha+1$ was also countable (since $\alpha+1=\alpha\cup\{\alpha\}$), and therefore a possible well ordering of $\omega$. This would contradict the above fact that $\alpha$ is greater or equal to all the ordinals which correspond to well-orderings of $\omega$, since $\alpha<\alpha+1$.
This means that $\alpha$ is uncountable, and that it is the first uncountable ordinal, since if $\beta<\alpha$ then $\beta$ can be injected into $\omega$, and so it is countable. Therefore we have that $\alpha=\omega_1=\aleph_1$.
Note that the above does not require the axiom of choice and holds in $\sf ZF$. The collection of all well-orders is a set by power set and replacement, so is the set of equivalence classes, from this we have that the collection of ordinals defined is also a set (replacement again), and finally $\alpha$ exists by the axiom of union. There was also no use of the axiom of choice because the only choice we had to do was of "a unique ordinal" which is a definable map (we can say when two orders are isomorphic, and when a set is an ordinal - without the axiom of choice).
With the axiom of choice this can be even easier:
From the axiom of choice we know that the continuum is bijectible with some ordinal. Let this order type be $\alpha$, now since the ordinals are well-ordered there exists some $\beta\le\alpha$ which is the least ordinal which cannot be injected into $\omega$ (that is there is no function whose domain is $\beta$, its range is $\omega$ and this function is injective).
From here the same argument as before, since $\gamma<\beta$ implies $\gamma$ is countable, $\beta$ is the first uncountable ordinal, that is $\omega_1$.
As to why there is no cardinals strictly between $\aleph_0$ and $\aleph_1$ (and between any two consecutive $\aleph$-numbers) also stems from this definition.
This is a function from the ordinals to the cardinals, and this function is strictly increasing and continuous. Its result is well-ordered, i.e. linearly ordered, and every subset has a minimal element.
This implies that $\aleph_1$ is the first $\aleph$ cardinal above $\aleph_0$, i.e. there are no others between them.
Without the axiom of choice, however, there are cardinals which are not $\aleph$-numbers, and it is consistent with $\sf ZF$ that $2^{\aleph_0}$ is not an $\aleph$ number at all, and yet there are not cardinals strictly between $\aleph_0$ and $2^{\aleph_0}$ - that is $\aleph_0$ has two distinct immediate successor cardinals.
For the second question, there is no actual limit. Within the confines of a specific model, the continuum is a constant, however using forcing we can blow up the continuum to be as big as we want.
This is the work of Paul Cohen. He showed that you can add $\omega_2$ many subsets of $\omega$ (that is $\aleph_2\le 2^{\aleph_0}$), and the proof is very simple to generalize to any higher cardinal.
In fact Easton's theorem shows that if $F$ is a function defined on regular cardinals, which has a very limited set of constraints, then there is a forcing extension where $F(\kappa) = 2^\kappa$, so we do not only violate $\sf CH$ but we violate $\sf GCH$ ($2^{\aleph_\alpha}=\aleph_{\alpha+1}$) in a very acute manner.