Zermelo's Well-Ordering Theorem/Converse/Proof 2

Proof
Let $\FF$ be an arbitrary collection of sets.

By assumption all sets can be well-ordered.

Hence the set $\bigcup \FF$ of all elements of sets contained in $\FF$ is well-ordered by some ordering $\leqslant$.

By definition, in a well-ordered set, every subset has a unique least element by $\leqslant$.

Also, note that each set in $\FF$ is a subset of $\bigcup \FF$.

Thus, we may define the choice function $C$:
 * $\forall X \in \FF: C: \FF \to \bigcup \FF$

by letting $\map C X$ be the least element of $X$ under $\leqslant$.