Choice function for powerset of $S$ entails choice function for $S$

Question

The Smullyan-Fitting book Set Theory and the Continuum Problem has the following exercise (Exercise 4.2):

Show that for any set $S$, if there exists a choice function for $\mathcal{P}(S)$, then there exists a choice function for $S$.

Assuming the Axiom of Choice, this is trivial. But could this be shown to hold without AC?

(Note: this is not the next exercise in the book, which has been shown impossible to prove without AC here, and various other places.)

EDIT: it seems that this question is ambiguous relative to which definition of a choice function we mean. The authors mean by 'choice function for $S$' the function $C$ with domain $\mathcal{P}(S)-\{\emptyset\}$ s.t. $C(x) \in x$. In this case, it is solvable by the accepted answer below.

Answer 1

I'm reading the question as follows:

Suppose there is a choice function for $\mathcal{P}(S)$: that is, a function $C:\mathcal{P}(\mathcal{P}(S)) \setminus \emptyset \rightarrow \mathcal{P}(S)$ such that for all non-empty $A \subset \mathcal{P}(S)$, $C(A) \in A$. Then there is a choice function for $S$, a function $D: \mathcal{P}(S) \setminus \emptyset \rightarrow S$ such that etc. etc...

If this is the correct interpretation of the question, the proof is as follows.

Let $C:\mathcal{P}(\mathcal{P}(S)) \setminus \emptyset \rightarrow \mathcal{P}(S)$ be the given choice function. We need to construct a choice function $D: \mathcal{P}(S) \setminus \emptyset \rightarrow S$ by working out how it should act on each non-empty $A \subset S$.

Consider the set $\bar{A} = \{ \{s\} : s \in A \}$. This is a non-empty set of subsets of $S$, so it is in the domain of $C$. Now consider $C(\bar{A})$, which is an element of $\bar{A}$ since $C$ is a choice function. All elements of $\bar{A}$ are singletons. Therefore we can define $D(A)$ to be the sole element of $C(\bar{A})$.

$D(A)$ is an element of $A$ because all of the singletons in $\bar{A}$ are singletons of elements of $A$. Therefore $D$ defined in this way is a choice function.

Answer 2

This is equivalent to "the axiom of choice from well-ordered families", that is, whenever the index set can be well-ordered, then there is a choice function.

As noted in the comments, if $\mathcal P(S)$ admits a choice, then $S$ can be well-ordered, but also vice versa is true.

However, this choice axiom is weaker than the full axiom of choice.