Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$.


For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$. 


If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $\[-n,n\]\cap A$ is non empty and then pick the infimum of this non empty compact set.


The question is the following: can you define a choice function for, say, $S=F\_{\sigma}\setminus \{\emptyset\}$ or $G\_{\delta}\setminus\{\emptyset\}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice? 



Choice function for Borel sets?
It is impossible without using the Axiom of Choice to prove the existence of such a choice function for $F\_{\sigma}$
sets, which include the countable sets. To see this, it is easier to work with the space $2^{\mathbb{N}}$ of infinite binary sequences instead of $\mathbb{R}$. Let $E$ be the eventual equality equivalence relation on $2^{\mathbb{N}}$. Then there is no measurable function which selects an element from each $E$-class.


