Can we get 'Transcendental Translates' without choice? This question asks for a proof that for any countable set $S\subset \mathbb{R}$ there's some number $c$ such that $(S+c)\cap\overline{\mathbb{Q}}=\emptyset$, where $S+c=\{s+c: s\in S\}$ is a translate of the set $S$; in other words, that every countable set has a translate all of whose members are transcendental. The 'natural' proof by cardinality arguments seems to implicitly use at least countable choice (countability of the algebraic closure of $S$ seems to require that the union of countably many countable sets is countable); is this necessary?
At first glance, the question seems very similar to the existence of a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, so a negative answer wouldn't be too surprising here; the structure of the algebraic closure (and in particular the well-ordering on polynomials) gives some amount of order to the problem, but without something like an explicit well-ordering on $S$ to order $\overline{\mathbb{Q}(S)}\cap\mathbb{R}$ I don't see that it's necessarily enough. So my core question is: assuming that AC doesn't hold, can there be a countable set $S$ such that every translate of $S$ includes an algebraic number? Going further still (I think), can we even have a countable set $S$ such that $\overline{\mathbb{Q}(S)}\cap\mathbb{R}=\mathbb{R}$?
 A: The countability of the algebraic closure is a choiceless proof, as long as you require that it is really a subset of $\Bbb C$, or even just linearly orderable (as a set, not a field!). And since we're talking about real numbers, this is fine.
The reason being is that the field is countable, so we can enumerate the polynomials, each have only finitely many roots, and using the linear order we can uniformly enumerate these roots. And using this we get uniform enumeration of the intermediate fields, and through that an enumeration of the algebraic closure (or the real closure, which is what you're actually interested in).
Finally, to make this an overkill proof, fix an enumeration of $S$, say $f$, and consider $L[f]$, which is a model of $\sf ZFC$ which contains $S$ as a set of reals. Note that any algebraic number over $\Bbb Q(S)$ is already in that inner model. Now apply the choice-y argument there and note that these translates are the same as in the "full universe" that contains all the real numbers.
