Zorn's Lemma implies the Axiom of Choice Can someone please give me feedback on my attempted proof that Zorn's Lemma implies the Axiom of Choice? I have a very good idea how to do it, but need help with some small details. 
This is my proof so far... major gaps missing, please help!
Let $X$ be a nonempty set.
Let $S = \{f : f \text{ is a choice function on } A \subset \mathcal{P}_{0}{X}\}$. Explain why $S$ nonempty. Declare $f_{1} \leq f_{2}$ if $Dom(f_{1}) \subset Dom(f_{2})$ (Why is this a partial ordering?).
Let $T$ be a chain in $S$. Let $F$ be the union on the functions in $T$. (Why is $F$ also a choice function?)
Therefore, by Zorn's Lemma, there exists a maximal function $m$. (Why is $m$ a choice function on every $A \subset PX$). Suppose not, then there exists an $A \subset \mathcal{P}_{0}{X}$ such that $m$ is not defined on $A$ (Don't know how to finish this!!)
Therefore, $f \in S$ and $m < f$, contradicting $m$ being maximal. Thus, $m$ is a choice function for the entire set $\mathcal{P}{X}$.
 A: Fixing your proof:
Let $X$ be a nonempty set.
Let $S = \{\,f : f \text{ is a choice function on } A \subset \mathcal{P}{X}\}$. 
$S$ is nonempty, as $f=\big\{(\{a\},a)\big\}\in S$, for all $a\in A$. Here, $\,f=\big\{(\{a\},a)\big\}$ is the choice function $\,f:\{\{a\}\}\to\{a\}$, with $\,f(\{a\})=a$. Note that in Set Theory, functions are identified with their graph, i.e. if $f:A\to B$, then $f=\{(a,f(a)):a\in A\}$.
Next, define the partial order: 
$$f_{1} \leq f_{2}\quad\text{iff}\quad f_{1}\subset f_{2}.
$$
It is a partial ordering, as "$\subset$" is always a partial ordering in $\mathcal PX$.
Let $T$ be a chain in $S$ and $F$ be the union on the functions in $T$. Then $F$ is a function (why?) and hence a choice function.
Therefore, by Zorn's Lemma, there exists a maximal function $f_{max}$. 
If the domain of $f_{max}$ is not the whole of $X$, then choose $a\in X\setminus \mathrm{Dom}\,f_{max}$, and set 
$$
\tilde f=f_{max}\cup\big\{(\mathrm{Dom}\,f_{max}\cup\{a\}),a\big\}.
$$
We have extended $f_{max}$ to $\tilde f$ by adding just one more element. Contradiction as $T\cup \{\tilde f\}$ would also be a chain, larger than $T$.
