For $$ an indexed function with nonempty terms, define a function $f$ such that $dom(f)=I$ and $(\forall i\in I)(f(i)\in A_i)$. Q: Let $<A_i:i\in I>$ an indexed function such that $A_i\not= \emptyset (\forall i\in I)$.  Since every set has a well-ordering, $\bigcup_{i\in I}A_i$ has a well ordering $≤$. Define a function $f$ such that $dom(f)=I$ and $(\forall i\in I)(f(i)\in A_i)$.  Thus, $<f(i):i\in I>$ is a choice function. A: By the definition of the axiom of choice, we have that since $<A_i:i\in I>$ is defined as such, there is an indexed function $<x_i:i\in I>$ such that $x_i\in A_i$ for all $i\in I$.  Is this not what we need?  The notation is starting to confuse me quite a bit.
A: As each $A_i$ is a non-empty subset of the well-ordered set $(\bigcup_{i \in I} A_i, \le)$, for each $i$, $\min(A_i)$ is well-defined unique element of $A_i$.
So we can define, without AC, $f(i)=\min(A_i)$ for each $i \in I$ and by definition this is a choice function for the family $\langle A_i: i \in I \rangle$.
For general indexed families of non-empty sets, we do need AC to guarantee the existence of a choice function for that family. The point being that we cannot make, in general, infinitely many arbitary choices. But having a well-order we can define an element in each subset in a unique way. So for well-ordered sets we don't need AC.  
It's like Hilbert's analogy: we do not need choice for a choice function for pairs of shoes (we can take the left shoe for each pair) but for pairs of identical socks we don't have such a definition (assuming we don't have left and right socks)..
A: It appears the problem is to prove the axiom of choice by assuming every set can be well ordered.  Your proof is just treading water, stating the conclusion to prove the conclusion.
Let A = $\bigcup_{i\in I}A_i.$
Well order A.
Define f:I -> A, i -> min $A_i \cap A.$
Show f is a choice problem.  
Indexing can be tedious.  Here is the same proof without indexing.
Let C be a collection of not empty sets.
Well order A = $\cup C.$
Show f:C -> A, X -> min $X \cap A$
is a choice function.
