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Consider the following partition of $\mathbb{Q}^2$ into disjoint classes : $(p_1, p_2) \sim (q_1,q_2)$ if and only if both $p_1 - q_1$ and $p_2 - q_2$ have odds denominators when written in their lowest terms.

Suppose I endow the equivalence class containing $(0,0)$ with some structure (say, a graph structure and a coloring of edges) and suppose I want to transfer this structure to all the other equivalence classes by translation. Do I need some weaker form of choice (like $\textsf{AC}_\omega$ for instance) to do this ?

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No. Since $\Bbb Q$ is provable countable, then $\Bbb Q^2$ is provable countable. Therefore we can define all the choice functions we need here, by fixing an enumeration of $\Bbb Q^2$ and considering the least element in the enumeration satisfying whatever we wanted.

(The same argument holds for any countable set. Note, however, that if your set does not have any definition which is provably countable, then it is going to be hard to define such an equivalence relation on your set.)

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    What does "provable countable" mean ? I presume it means that there is a known explicit complete enumeration of $\mathbb{Q}$. Am I right ? – 2017-01-22
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    So if we replace $\mathbb{Q}$ by some countable set which is not provable countable (assuming such a thing sense; I obviously don't know what I'm talking about), some weaker form of choice would be necessary, right ? – 2017-01-22
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    Well, if something is countable, then it is countable in $V$, and the same arguments hold. if you replace $\Bbb Q$ by a definition for a set which may or may not be countable or even well-orderable, then how do you even define the equivalence relation? – 2017-01-22
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    You are right. Silly question. The setting I described only make sense for $\mathbb{Q}$. I was just trying to see how the situation could be so different that some weak for of choice would be needed. – 2017-01-22
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    So I do not need $\textsf{AC}_\omega$ to arbitrarily select a single element from each set of a countable collection of subsets of a given countable set. But I do need it to arbitrary select a single element from each set of a countable collection of countable sets in general. Is this correct ? – 2017-01-22
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    Yes, that is correct. – 2017-01-22