Is the axiom of choice necessary to refer to a set $S$ if we already know that $S$ exists? Consider the equivalence relation on $[0,1]$ where
$$x \sim y \iff x - y \in \mathbb{Q}.$$
Fix a set $S$ which contains exactly one representative from each equivalence class of the relation. It is well-known that constructing such a set requires the axiom of choice. However, I don't have an intuition for what breaks down if the axiom of choice is not assumed.
If I don't assume the axiom of choice, what happens to $S$? Does it cease to exist? If not, then surely $S$ is some subset of $[0,1]$, whether or not we invoke the axiom of choice, so why can't I refer to $S$ in a proof without choice, in the same way that I would refer to any other subset of $S$?
I haven't studied set theory formally, so please excuse my ignorance.
 A: Nothing "happens" to $S$; but you might simply not be able to prove such an $S$ exists.
Note your description of $S$ : it's a set that contains one representative of each equivalence class. Why should it exist ?
Saying "Fix such a set $S$" requires a proof that it exists.
Let me start with an easier example: suppose you're working with some set of numbers, but you don't know whether it's $\mathbb R$ or $\mathbb C$. You're also given some $x$, and you want to say "fix a $y$ be such that $y^2=x$" : why does it exist ? If you know you're working in $\mathbb C$ (analogous to assuming the axiom of choice), then there's a theorem (analogous to the axiom of choice) which tells you that such a $y$ exists, so you're well and good.
But if you don't know and might actually be working in $\mathbb R$ (analogous to being in a situation where AC fails), you can't be sure that such a $y$ exists, and so you can't prove it; and so you can't "use" the $y$ (of course you can still prove properties about such $y$'s, just as here you can prove things on such $S$'s, but that tells you nothing about whether they exist !)
Back to your situation: one point of view on the statement that AC cannot be proved from ZF (assuming ZF is consistent) is that there are "universes of math" where AC simply does not hold. In such universes, it would be possible that no such $S$ exists. It didn't "cease" to exist, it simply never did. But there are other "universes of math" where such an $S$ might exist.
If you could give an explicit definition of $S$, then you could always refer to it (although, it might be non obvious why this definition actually defines a set).
