Showing that two spaces are isomorphic Let $\sim$ be the equivalence relation on $\mathbb{R}$ defined by: for all x,y in $\mathbb{R}$:
$ x\sim y$ iff $x - y \in \mathbb{Q}$
I also have an equivalence relation at $\mathbb{R^2}$ defined by: ($x_0$, $x_1$) $\sim_2$ ($y_0$, $y_1$) if and only if $x_0 - y_0 \in \mathbb{Q}$ and $x_1 - y_1 \in \mathbb{Q}$
Proof that the structure ($\mathbb{R}$,$\sim$) is isomorphic with ($\mathbb{R^2}$,$\sim_2$)
I already know in both structures I have countable many elements per equivalence class, and there are uncountably many, the same cardinality as $\mathbb{R}$, of that classes. So I wanted to make such an isomorphism that sends classes to other classes, but I don't know how to do it.
Is it sufficient to use the axiom of choice uncountable many times to choose a new class every time? 
 A: Constructing an explicit bijection is going to be hard and not particularly enlightening. The intended solution is most probably more abstract than that.
Do you already know that $|A\times B|=\max(|A|,|B|)$ when at least one of $A$ and $B$ is infinite?
Can you prove that each equivalence class (in both structures) is countably infinite?
Since $|\mathbb R|=|\mathbb R^2|$, what does this say about the number of equivalence classes in each structure? (Hint: not merely that there are "uncountably many").
If you can show that there are equally many equivalence classes, what you have proved is exactly that there is some bijection between one set of equivalence classes and the other, no matter that you may not have a nice way to define that bijection.
A: One way is to start with a bijection $\varphi:\Bbb Q\to\Bbb Q^2$. $\Bbb Q$ is one of the $\sim_1$-equivalence classes, and $\Bbb Q^2$ is one of the $\sim_2$-equivalence classes, so this is at least a start. Now we want to extend $\varphi$ to a bijection $\widehat\varphi:\Bbb R\to\Bbb R^2$ in such a way that for all $x,y\in\Bbb R$, $x\sim_1 y$ iff $\widehat\varphi(x)\sim_2\widehat\varphi(y)$.
One nice way is to appeal to the axiom of choice twice to get a set $A\subseteq\Bbb R$ that contains exactly one member of each $\sim_1$-class except the class $\Bbb Q$ and a set $B\subseteq\Bbb R^2$ that contains exactly one member of each $\sim_2$-class except the class $\Bbb Q^2$. $|A|=|B|=|\Bbb R|$ (why?), so there is a bijection $\psi:A\to B$. Now let 
$$\widehat\varphi:\Bbb R\to\Bbb R^2:x\mapsto\begin{cases}
\varphi(x),&\text{if }x\in\Bbb Q\\
\psi(a)+\varphi(x-a),&\text{if }x\sim_1 a\in A\;,
\end{cases}$$
and prove that $\widehat\varphi$ has the desired properties. (Addition in $\Bbb R^2$ is, as usual, component-wise.)
