Here is Prob. 5, Sec. 10, in the book Topology by James R. Munkres, 2nd edition:
Show the well-ordering theorem implies the choice axiom.
Here is the Well-Ordering Theorem:
If $A$ is a set, there exists an order relation on $A$ that is a well-ordering.
And, here is the Axiom of Choice:
Given a collection $\mathscr{A}$ of disjoint non-empty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathscr{A}$; that is, a set $C$ such that $C$ is contained in the union of the elements of $\mathscr{A}$, and for each $A \in \mathscr{A}$, the set $C \cap A$ contains a single element.
Now here is my attempt at Prob. 5, Sec. 10:
We assume that the well-ordering theorem holds.
Let $\mathscr{A}$ be a (non-empty) collection of disjoint non-empty sets.
Then, for each $A \in \mathscr{A}$, there exists an order relation $<_A$ on $A$ that is a well-ordering, and under this relation, the set $A$, being a non-empty subset of itself, has a smallest element, which we call $a_A$.
Let us form the set $C$ as follows: $$ C = \left\{ \ a_A \ \colon \ A \in \mathscr{A} \ \right\}. $$ Then $C$ is a set that has exactly one element from each set in $\mathscr{A}$.
Is this proof correct? If not, then where are problems in it?
Does the converse of this result hold too? If so, then how to prove it? If not, then how to construct a counter-example?
Alas, this attempt assumes the axiom of choice. In effect one chooses a well-ordering for each $A\in\scr A$. To do this one needs AC. To avoid this, let $U=\bigcup_{A\in\scr A}A$ and take a well-ordering on $U$. Each $A\in\scr A$ is a subset of $U$, so has a first element with respect to the well-ordering of $U$.