First of all, I suggest you find a good article/book on Cantor's work on multiple infinities. Reading that will enlighten you and answer probably most of the questions you have concerning this subject. Try these lecture notes.
But to give a brief answer to your current questions:
Per definition, two sets have the same cardinality if there exists a bijection between them. This definition was introduced by Cantor because for infinite sets, you could not just count the elements. Not all infinite sets have the same cardinality. For example, the natural numbers $\mathbb{N}$ and the real numbers $\mathbb{R}$ do not have the same cardinality. This was proven with the diagonal argument.
$\mathbb{N}$ and $\mathbb{Z}$ are both infinite sets. I suggest you check out their definitions on wikipedia (the natural numbers, the integers). They also, like Ali said in his post, have the same cardinality. The bijection that was given by him is probably the easiest one to grasp.
As for $\mathbb{Q}$, the rational numbers, this is also an infinite set that has the same cardinality as $\mathbb{N}$ (and thus also $\mathbb{Z}$). I suggest you check out the bijection that is given in the lectures notes that I linked above, that one is probably the clearest one.
Under the axiom of choice, for any two infinite cardinal numbers, their sum, which corresponds to their disjoint union, is simply the maximum of the two numbers. This is a standard fact of cardinal arithmetic you can find in every set theory book.
So if $S$ is uncountable with cardinal $\alpha$, we have $\alpha+\alpha=\alpha$. So their exists two disjoint copies $S_1, S_2$ of $S$ such that their union has the same cardinality as $S$. So there is a bijection $f:S_1\cup S_2\to S$. So $f(S_1)$ and $f(S_2)$ are two disjoint uncountable subsets of $S$.
Here is an actual method to "construct" a bijection from $S$ to $S\times\{0,1\}$. Let $\mathcal{F}$ be the family of all bijections of some subset $T$ of $S$ to $T\times\{0,1\}$. Such a bijection always exists when $T$ is a countably infinite subset. Order these functions by set inclusion on their graph. Then the conditions of Zorn's Lemma are satisfied, and their exists a maximal such function of the form $f:T\to T\times\{0,1\}$. If $S\backslash T$ would be infinite, it would contain a countable subset and then we could extend $f$ to a larger such function in contradiction to it being maximal. So $S\backslash T$ must be finite. So there is a bijection $g:\{0,1,\ldots,n-1\}\to S\backslash T\times\{0,1\}$. Let $C=\{x_0,x_1,\ldots\}\subseteq S$ be countably infinite. You can construct now a new bijection $f':S\to S\times\{0,1\}$ by
$$f'(x) = \begin{cases} g(m) &\mbox{if } x_m\in\{x_0,\ldots,x_{n-1}\}\subseteq C\\
f(x_{m-n}) & \mbox{if } x_m\in C\text{ and }m\geq n.\\
f(x) &\mbox{otherwise.}\end{cases} $$
The proof is essentially from Halmos little set theory book.
Best Answer
You've asked two questions; I'll answer them separately.
Question 1: if two sets are of the same cardinality, are they equinumerous?
The answer is yes, by definition. However, you ask in the comments about how to prove this, so I'll elaborate more on the definitions. It's been pointed out in the comments that two sets are equinumerous if there is a bijection between them. When we say two sets have the same cardinality, we are using one of two definitions.
Usually, when we say two sets have the same cardinality, we are talking about cardinal numbers, a special kind of ordinal number (which you can look up on wikipedia). If we assume the axiom of choice, then every set $A$ is in bijection with a unique cardinal number, denoted $|A|$, which we call its cardinality. Examples include $\aleph_0$, the cardinality of countable sets such as $\mathbb{N}$ or $\mathbb{Z}$, and $2^{\aleph_0}$, the cardinality of $\mathbb{R}$. We can then say things like "$\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality", because $|\mathbb{N}| = \aleph_0 = |\mathbb{Z}|$.
However, one might also just say that two sets "have the same cardinality" if they are equinumerous; that is, if there is a bijection between them. This is a definition more commonly used in beginner math courses, when you don't want to get into the details of cardinal numbers or the axiom of choice. (This second definition follows almost immediately from the first: if $A$ and $B$ both have cardinality $\alpha$, there are bijections $f : A \rightarrow \alpha$ and $g : B \rightarrow \alpha$, and so $g^{-1} \circ f : A \rightarrow B$ is a bijection too, meaning $A,B$ are equinumerous.)
Hopefully, this convinces you that "$A$ and $B$ have the same cardinality" implies "$A$ and $B$ are equinumerous", if not by definition, then almost immediately.
Question 2: are any two uncountable subsets of $\mathbb{R}$ equinumerous?
The answer to this is: it depends on the continuum hypothesis. The continuum hypothesis states that $2^{\aleph_0}$ (the cardinality of $\mathbb{R}$) is the next biggest infinity after $\aleph_0$ - there are no other sizes of infinity in between. (This is often written $\aleph_1 = 2^{\aleph_0}$.) Unfortunately, the continuum hypothesis has been shown to be independent of the ZFC axioms, so your question has no definitive answer; but, we can still consider the two cases.
If the continuum hypothesis is true, then the answer to your question is yes. Indeed, if a subset $A$ of $\mathbb{R}$ is uncountable, then we have $\aleph_0 < |A| \leq |\mathbb{R}| = 2^{\aleph_0}$, and the continuum hypothesis then implies that we must have $|A| = 2^{\aleph_0}$ ($= |\mathbb{R}|$). Thus $|A|$ must be equinumerous to $|\mathbb{R}|$. On the other hand, if the continuum hypothesis is false, then we could find an uncountable subset of $\mathbb{R}$ whose cardinality is smaller than the cardinality of $\mathbb{R}$, so the answer to your question would be no.
(There's a slight issue here because we're assuming $A$ does have a cardinality. Without the axiom of choice, I think the continuum hypothesis could be true while at the same time there exists a subset of $\mathbb{R}$ which doesn't have a cardinality, hence is not equinumerous to $\mathbb{R}$. But that's a bit beyond me; usually when talking about the continuum hypothesis, the axiom of choice is assumed.)