Proof of Choice for a Family Indexed by a Finite Set

If $F=\emptyset$, then the empty function $a:\emptyset\to S$ satisfies the required condition vacuously, since there is no $i\in F$.

Suppose from now on that $F\ne\emptyset$. Since $F$ is finite and nonempty, there is a natural number $n$ such that $F$ has $n$ elements, that is, such that there is a bijection $f:[n]\to F$, where $[n]$ denotes the initial segment determined by $n$. By claim 1 of Properties of the Order on the Natural Numbers one has $n\le n$, so $n\in[n]$ and $f(n)$ is defined.

Define a family of subsets of $S$ $(B_m)_{m\in\mathbb{N}}$ indexed by $\mathbb{N}$ by

This is well defined: if $m\le n$ then $m\in[n]$, so $f(m)\in F$ and $A_{f(m)}$ is one of the given sets, and $f(n)\in F$ in the remaining case. In particular every $B_m$ equals $A_i$ for some $i\in F$ and is therefore a nonempty subset of $S$.

By Axiom of Countable Choice there exists a sequence $(b_m)_{m\in\mathbb{N}}$ in $S$ such that $b_m\in B_m$ for every $m\in\mathbb{N}$.

Define $a:F\to S$ as follows. Let $i\in F$. Since $f$ is a bijection, there is exactly one $k\in[n]$ with $f(k)=i$; put $a(i)=b_k$. The uniqueness of $k$ makes this assignment well defined, and no choice is involved, because $k$ is determined by $i$.

Finally, fix $i\in F$ and let $k\in[n]$ be the unique index with $f(k)=i$. Then $k\le n$, so $B_k=A_{f(k)}=A_i$, and therefore

which is the required conclusion.