You can indeed reduce AC to "AC for well-ordered processes" - however, AWC is "AC for well-ordered index sets," which is quite a different (and weaker) thing.
To see the difference, let's follow your idea for a two-element set $X=\{a,b\}$. Our first "stage" is: pick an element $e_0$ from the set $X$. Our second stage is: pick an element $e_1$ from the set $X\setminus\{e_1\}$.
On the face of it this may seem like we've used the (trivial) axiom of choice for two sets. However, we haven't really done that, since our second set depended on our first choice. For AWC you need to set forth ahead of time a well-ordered set of choices you want to make. But you need a bit more than that to prove WOP: basically, you need "well-ordered dependent choice" rather than simply well-ordered choice. And the problem is that while the set of "stages" in your construction is well-orderable, the set of sets you might need to choose from isn't (at least, a priori), and so mere AWC won't help you here.
Incidentally, note that DC - which is basically "dependent choice for $\omega$-many stages" - is indeed strictly stronger than the axiom of countable choice, which is the "index-set-instead-of-stages" version. So there's a pattern here.
After some hard work, I found the following example:
Zorn's lemma implies the Zermelo well-ordering principle: Let $S$ be a set. We will prove that there exists a relation in $S$ that is a well-order of the set. Let:
$$\mathcal{W}=\{\langle A,<_A\rangle|A\subset S\;\&\;<_A\subseteq A\times A\text{ is a well-ordering of }A\}$$
Clearly, $\mathcal{W}\not=\emptyset$; in fact, from the Russell-Whitehead theorem we know that every finite subset $A\subseteq S$ can be well-ordered. We will define in $\mathcal{W}$ the following relation:
$$\langle A,<_A\rangle\preccurlyeq\langle B,<_B\rangle\iff\langle A,<_A\rangle\text{ is an initial segment of }\langle B,<_B\rangle$$
Or in other words:
- $A\subseteq B$
- $<_A\subseteq <_B\cap(B\times B)$
- For each $a\in A$ and $b\in B\setminus A$, we have that $a<_B b$
Clearly, $\prec$ is irreflexive and transitive, so $\langle\mathcal{W},\prec\rangle$ is a partially ordered set.
Let $C$ be a chain of $\langle\mathcal{W},\prec\rangle$. We will see that $C$ has an upper bound in $\langle\mathcal{W},\prec\rangle$. Let $U=\bigcup \text{dom}(C)$, and for each $s,t\in U$, we will say that $s<_U t$ if $s,t\in A$ and $s<_A t$ for some $\langle A,<_A\rangle$.
$<_U$ is then clearly well-defined and is a linear ordering of $U$. If $A\subset U$, then there exists $\langle B,<_B\rangle\in C$ such that $A\cap B\not=\emptyset$, and therefore, $A$ admits a minimal element $a_B$ (in the sense of $<_B$).
We will see that this can be defined to be the minimal element of $A$ in the sense of $<_U$. If it is also true that $a\cap B'\not=\emptyset$, for another set $B'$ with $\langle B',<_{B'}\rangle\in C$, since $C$ is a chain of $\langle W,\prec \rangle$, $\langle B,<_B\rangle\preccurlyeq\langle B',<_{B'}\rangle\text{ or }\langle B',<_{B'}\rangle\prec\langle B,<_B\rangle$. For example, suppose that $\langle B,<_B\rangle\preccurlyeq\langle B',<_{B'}\rangle$ (the other case is analogous). This implies that the minimal elements $a_B$ and $a_{B'}$ of $A\cap B$ and $A\cap B'$ are the same, since $\langle B, <_B\rangle$ is an initial segment of $\langle B',<_{B'}\rangle$, or $B=B'$.
In the end, $\langle U,<_U\rangle\in\mathcal{W}$ and it is an upper bound of $C$ in $\langle\mathcal{W},\prec\rangle$.
By Zorn's lemma, $\mathcal{W}$ admits a maximal element $\langle M,<_M\rangle$ in the sense of $\prec$. Lets see that, in fact, $<_M$ is a well-ordering of the set $S$. On the one hand, it is clear that $<_M$ is a well-order of a subset of $S$. On the other hand, if there existed an element $s\in S\setminus M$, then we could consider the set $M'=M\cup \{s\}$ and the relation $<_{M'}$ (that extends $<_M$) defined by: for each $t\in M$, $t<_{M'} s$. But then $\langle M', <_{M'}\rangle\in\mathcal{W}$, contradicting the fact that $\langle M,<_M\rangle$ is maximal.
Therefore, $<_M$ is a well-order of the set $S$.
Best Answer
There are two key observations here:
A chain that contains its own upper bounds either contains a maximal element, or does not have a strict upper bound.
Zorn's lemma is equivalent to Zorn's lemma where we require that the chains are well-ordered.
The first observation is quite trivial. The second, however, lies within the standard proof of Zorn's lemma. Think about it.
How do we prove Zorn's lemma? We start with a partial order, and we construct a well-ordered chain by choosing, one step at a time, an upper bound. Then we argue, we had to have come to a stop, which provides us with a maximal element.
So to prove Zorn's lemma we only need to focus on well-ordered chains.
Your root mistake is, I think, misunderstanding the meaning of "well-ordered" here. When we say, for example, that $A$ is a well-ordered subset of $\Bbb R$, we usually mean by the standard order of the reals (which therefore implies $A$ is countable). So here when we say well-ordered chain in the partial order, we mean that the inherited order is indeed a well-ordering, and not just a linear ordering.
I mean, if you allow any well-ordering of the chain, then what is the point of having a partial order underlying the discussion to begin with? The question starts to lose meaning. And to some extent, is trivial now, since every well-ordered set, indeed any linearly ordered set, contains all of its upper bounds in itself.