Zorn's Lemma implies Axiom of Choice

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Let $S$ be a collection of nonempty sets and $F$ be the collection of all functions $f$ for which ${\rm dom}(f) \subseteq S$ and $f(X)\in X$ for all $X \in {\rm dom}(f)$. The set $F$ is ordered by inclusion $\subseteq$.

Assume that $C$ is a chain in $(F,\subseteq)$. Let $f_0=\bigcup C$. It is easy to verify that $f_0 \in F$ and $f_0$ is an upper bound of $C$. Then $F$ has a maximal element $\bar f$ by Zorn's Lemma. To accomplish the proof, we next show that ${\rm dom}(\bar f)=S$. If not, ${\rm dom}(\bar f) \subsetneq S$ and thus $S \setminus {\rm dom}(\bar f) \neq \emptyset$. Take some $\bar X \in S \setminus {\rm dom}(\bar f)$ and $\bar x\in \bar X$. Clearly, $\bar f \bigcup \{(\bar X,\bar x)\}\in F$. This contradicts the minimality of $\bar f$. This completes the proof.


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