Proving Zermelo's Theorem implies the Axiom of Choice I thought it would be fun to try and prove this. It turned out to be pretty simply and maybe too simple so I was wondering if anybody could verify if this proof is correct. 
Suppose we have a family $\mathbb{F}$ of nonempty sets. By Zermelo's Theorem each one has a least element and this constructs a choice function $f$ such that $f(A) \in A$ is the least element of $A$ for all $A \in F$.
 A: Sadly, it isn't quite that easy, because then you have to choose a well-ordering for each set in the family. Instead, try well-ordering $\bigcup\Bbb F$, and see where that gets you.
A: Indeed Cameron is right, you used the axiom of choice in choosing a well-ordering for each $A$ in the family; and if you instead well-order $\bigcup\Bbb F$ then you can apply the same argument.
There is another method, though, and that is to well-order the set of finite choice functions. That is consider the set, $$\mathscr C=\left\{\prod_{i=1}^n A_i\mathrel{}\middle|\mathrel{} A_i\in\Bbb F, n\in\omega\right\}$$
This set is clearly not empty, and fix a well-ordering $\leq$ on $\scr C$. Now define the choice function to be as follows: $f(A)=a$ if and only if the $\leq$-minimal $g\in\scr C$ such that $A\in\operatorname{dom}(g)$ and $g(A)=a$. It is not hard to see that such $g$ exists, either. So the function $f$ is well-defined and of course it is a choice function.
However, this method is less immediate and natural than the one suggested by Cameron.
