To show that Axiom of Choice $\Leftrightarrow$ Well-ordering Theorem,
Hrbacek and Jech in their book prove the following theorem first:
$X$ can be well-ordered if and only if $\mathcal{P}(X)$ has a choice function.
However, I fell I'm missing something. Sure, assuming the aforementioned theorem it's easy to prove that the axiom of choice implies the well-ordering theorem (let $X$ be a set, if every set has a choice function, so does $\mathcal{P}(X)$, thus, $X$ can be well-ordered).
However, how to deduce the converse using the theorem?
Assume that every set can be well-ordered. Let $X$ be a set. To use the theorem, we need to consider $X$ as a power set for some other set $A$? However, I'm not sure that every set is a power set of some other set. In fact, a nonempty set need to contain $\varnothing$ in order to be a power set.
Best Answer
Given a set $X$, let $Y=\bigcup X$. Then $X\subseteq\mathcal{P}(Y)$, so a choice function on $\mathcal{P}(Y)$ can be restricted to give a choice function on $X$.
(Incidentally, as I would define it, you can only have a choice function on a set of nonempty sets, since you can't choose an element from the empty set. So really we should speak of choice functions on $\mathcal{P}(X)\setminus\{\emptyset\}$, not on $\mathcal{P}(X)$.)