In terms of purely set theory, the axiom of choice says that for any set $A$, its power set (with empty set removed) has a choice function, i.e. there exists a function $f\colon \mathcal{P}^*(A)\rightarrow A$ such that for any subset $S$ of $A$, $f(S)\in S.$ Is this correct?

My question then is about *proving this fact*, so that we do not need to put it as an axiom. Now as per the research done on this single object- *Axiom of Choice*, I believe here that there should be some falsity in my argument. I do not find the mistake.

For any $S\in \mathcal{P}^*(A)$, since $S\neq \emptyset$, $\exists s\in S$. Define $f(S)=s$. Then $f$ is a choice function.

This was showing me that *the axiom of choice is proved*, but then why it had been put as an axiom? For example, in this book, the author asserts that

It is a metatheorem of mathematical logic that it is impossible to specify the function that assigns to each non-empty subset of $\mathbb{R}$, an element of itself.

There are several notes and books on axiom of choice, but here I am trying to understand through *doing some argument for some problem*, where problem actually arises.