Zorn's Lemma implies Axiom of Choice

Theorem
If Zorn's Lemma is true, then so must the Axiom of Choice be.

Proof
Let $X$ be a set.

Let $\FF$ be the set of mappings defined as:
 * $f \in \FF \iff \begin{cases}

\Dom f \subseteq \powerset X & \ \\ \Img f \subseteq X & \ \\ \forall A \in \Dom f: \map f A \in A & \ \end{cases}$

Let $\preceq$ be the relation defined on $\FF$ as:


 * $\forall f_1, f_2 \in \FF: f_1 \preceq f_2 \iff f_2$ is an extension of $f_1$.

Straightforwardly, $\preceq$ is a partial ordering on $\FF$.

Suppose Zorn's Lemma holds.

Then there exists a maximal element of $\FF$.

We then show that if $g$ is such a maximal element, then:
 * $\Dom g = \powerset X \setminus \O$