We want to call $x$ and $y$ equivalent iff $x - y \in \Bbb Q$ for $x, y \in [0, 1].$ We choose exactly one representative from every equivalence class. Let $V \subset [0, 1]$ be the set of every representative that we chose before.
Now let $\{q_1, q_2, \ ... \}$ be an enumeration of every rational number in $[-1, 1]$. Show that the sets
$V_k := q_k + V := \{q_k + v: v \in V \}, k \in \Bbb N$
are disjoint to each other.
Now, I'd guess that there can be only two types of equivalence classes here and thus, only two types of representatives. On the hand, we have the equivalence class that contains all the rational numbers in $[0, 1]$. That is clear since $\Bbb Q$ is a field, therefore, every subtraction of two rational numbers in $[0, 1]$ gives us a rational number again. On the other hand, we have various equivalence classes that contain irrational numbers.
So, in order to show that the sets are disjoint to each other, we have to prove that $q_i + V \neq q_j + V$ for arbitrary $i, j \in \Bbb N, i \neq j$ respectively $q_i + v \neq q_j + v$ for arbitrary $v \in V$. What bothers me is that this seems to be rather trivial, right? $V$ just contains representatives who are rational or irrational numbers, the $q_k$ are rational numbers too. So the equation seems to be self-explaining. Or could there exists flaws when it comes down to comparing the translation of a rational vector and a irrational vector?
Suppose $V_k\cap V_j\neq \emptyset$ for $k\neq j$. Let $x\in V_k\cap V_j$. Then $x=q_k+v_k=q_j+v_j$ for some $v_k,v_j\in V$.
Thus, we have that $v_j-v_k=q_k-q_j\in\mathbb{Q}$. But this would then imply that $v_k\sim v_j$ under the equivalence relation. Since $V$ contains only $1$ representative from each class, we have to have that $v_j=v_k$. Then this must imply that $q_j=q_k$. But $j\neq k\Rightarrow q_j\neq q_k$. Thus $v_j\neq v_k$, so $x\neq x$, contradiction. So $V_j\cap V_k = \emptyset$.