Let $X\neq \emptyset$ be a set. We say ${\cal C} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ is a *cover* if $\bigcup {\cal C} = X$. A subset $D\subseteq X$ is a *choice set* for ${\cal C}$ if $|D\cap c| = 1$ for all $c\in C$.


[Ramiro de la Vega](https://mathoverflow.net/users/17836/ramiro-de-la-vega) showed in his [nice post](https://mathoverflow.net/a/285702/8628) that



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He used the Axiom of Choice to well-order the members of the cover and construct a subcover with a choice set.


**Question.** Does the statement (S) above imply the Axiom of Choice?



Choice sets and the axiom of choice
Let $P\subseteq X\times Y$ be sets such that $\forall x\in X\exists y\in Y[(x,y)\in P]$. We wish to derive from (S) the existence of an $f\subseteq P$ such that $\forall x\in X\exists! y\in Y[(x,y)\in f]$.


Consider the following cover of $P$. Let $
\cal{C}$ $:= \{\{(x,y)|y\in Y, (x,y)\in P\}|x\in X\}$. Notice that any subcover of $\cal{C}$ has to be $\cal{C}$ itself.


Let $f$ be a choice set for $\cal{C}$, then trivially $f\subseteq P$ and $\forall x\in X\exists! y\in Y[(x,y)\in f]$.


