Do We Need the Axiom of Choice for Finite Sets?

The ability to give a name to some arbitrary element of a nonempty set is known as "existential instantiation". This is an inference rule of the underlying logic, not part of the theory, and so it is included not only in ZFC, but also in ZF, and in appropriate forms it is also included in PA and every other first-order theory.

The intuition is that giving a name to an object does not really require "choosing" an object; if we know that there is an element in a set $A$, and we begin arguing about some "given" element $c \in A$, we do not have to actually make a choice, we can just reason hypothetically ("what if we knew an element $c$ of $A$?"). This reasoning will be sound because $A$ really is nonempty. However, because we don't know which element is chosen, the only properties we know it has are the ones we can prove every element of $A$ has. In particular, we cannot assume that the element we are given is equal to any other object we already have, unless we can prove they are equal. So we need to give a new name that we are not already using for any other object.

The word "choose" is used in several ways in set theory, both in this way (existential instantiation), and in ways corresponding to the axiom of choice, and in ways where there is not really any choice, for example when we have a function $f$ and a set $x$ and we "choose" $y$ such that $f(x) = y$. You have to look at the context to tell what formal proof is being suggested by the informal proof that you read.

You might be thinking that "choosing" a single element from a set would be an algorithmic operation, but this is just not the case. The nonconstructivity of existential instantiation is a direct consequence of the nonconstructive meaning of $\exists$ in classical logic. To get a system where you can effectively choose a witness to any true existential statement, you have to move to constructive mathematics, but then the meaning of $\exists$ changes as well.

Saying S is not empty, therefore there is an x∈S, does not suffice.

Ah, but it does suffice! Knowing that $S$ is non-empty, there do exist such $x$, and so we can declare the symbol $x$ to denote an element of $S$, and the rest of our argument is then a function of $x$.

To put it more formally, a typical proof that chooses an element from $S$ has the form

Let x∈S
...
Therefore P

and constitutes a proof of $\forall x: (x \in S \implies P)$, which in turn implies $P \vee (S = \emptyset)$.

Similarly, if you made two choices $x \in S, y \in T$, you typically prove a statement like $\forall x,y: \left( (x,y) \in S \times T\implies P \right) $.

This makes it clear how the axiom of choice enters the picture; a sequence of infinitely many choices indexed by $I$ of the form described above would be a proof of

$$ \forall x : \left(x \in \prod_{i \in I} S_i \implies P \right)$$

which implies

$$ P \vee \left(\prod_{i \in I} S_i = \emptyset \right) $$

i.e. a proof that $P$ is true or the family $S_i$ does not have a choice function (or both).

There are two things. Let $(A_i)_{i\in I}$ be an arbitrary family of nonempty sets. One version of the axiom of choice says that $\prod_i A_i\neq\emptyset$. We cannot prove this without the axiom of choice if we know that for all $i\in I$ the set $A_i$ is finite. What we can show is that the product is nonempty if $I$ is finite!

The proof goes like this: We show that for each $n$, if $I$ has $n$ elements, then the product is nonempty. We do this by induction. Without loss of generality, we let $I=\{1,\ldots,n\}$. If $I$ contains a single element, showing that the product is nonempty amounts to finding an element in the single set $A_1$. That this can be done is simply the meaning of $A_1$ being nonempty. So suppose there exists a choice function $(a_1,\ldots,a_{n-1})$ defined on $I\backslash\{n\}=\{1,\ldots,n-1\}$. Since $A_n$ is nonempty, there is some $a_n\in A_n$. This gives a choice function for $I$ given by $(a_1,\ldots,a_{n-1},a_n)$. And for this, we don't need the axiom of choice.