Show that a set is not countable using diagonalization argument 
We define the following equivalence relation on $\mathbb{R}$: $a \equiv_\mathbb{Q} b$ if $a-b \in\mathbb{Q}$. Let $A \triangleq \mathbb{R}/\equiv_\mathbb{Q}$ be the quotient set.
Use diagonalization to show that $A$ is not countable.

We can take all the representative from the interval $[0,1]$: different representatives are all the unique irrational numbers in $[0,1]$ plus $[0]=\mathbb{Q}$.
I need to show that for every $f:\mathbb{N}\to A$ exists some $[a]\in A\setminus f(\mathbb{N})$. It is suffice to find some $a\in\mathbb{R}$ s.t. $a \not\equiv_\mathbb{Q} r_n$ for all $n\in\mathbb{N}$ where $[r_n]=f(n)$. How does one go about to "construct" such an $a$?
Hint?
 A: This question could likely only be answered correctly by the person who assigned it, and who has a particular "correct" solution in mind. 
Here is my proposal (a matter of taste perhaps, regarding how exactly it is done). List the rationals $\mathbb Q$ as $\{q_1,q_2,q_3,\dots\}$ and let $Q_n =  \{q_1,q_2,\dots, q_n\}$. Then  each $Q_n$ is finite, and $\mathbb Q$ is the increasing union of the $Q_n$. For each $x$ let $x+Q_n= \{x+q:q\in Q_n\}$. Clearly $x+Q_n$ is finite for each $x$ and each $n$. 
Given $f:\mathbb{N}\to A$, pick a representative $x_n\in f(n)$. Let $I_0=[0,1]$ and recursively pick non-degenerate closed intervals $I_n$, $n\ge1$, such that $I_{n+1}\subseteq I_n$ (for all $n\ge0$), and $I_n\cap(\bigcup_{1\le k\le n}(x_k+Q_n) =\emptyset$. This is possible since $\bigcup_{1\le k\le n}(x_k+Q_n)$ is finite, for each $n$. Then the sequence of closed intervals $I_n$ is decreasing (and each interval is bounded, and we could also have asked that the lenght of $I_n$ is at most $\frac1n$, for convenience), hence there is an $x\in\bigcap_{n\ge1}I_n$. Then $[x]\in A\setminus f(\mathbb N)$, which completes the proof. 
