Axiom of Choice is equivalent to Well-ordering Theorem: Hrbacek, Jech - "Introduction to Set Theory"

Question

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.

Answer 1

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 $\mathcal{P}(X)$.)

Answer 2

I presume when they say $P(X)$ has a choice function, they mean that there is $F:P(X)\setminus\{\emptyset\}\to X$ with $F(A)\in A$ for all nonempty $A\subseteq X$? If there is such a choice function, transfinite induction gives a well-ordering of $X$.

The existence of such a choice function for all $X$ is clearly equivalent to AC.