The theorem that every partial order with well ordered chains must be well founded can be proved with the Hausdorff Maximal Principle which is equivalent to the axiom of choice. Does this theorem imply the axiom of choice? The proof I have thought was that given a set of sets $A$ we define on it's union a partial order where $a< b$ if a is an element of the transitive closure of b. This can be verified to be a partial order since transitivity follows from the definition of transitive closure and irreflexivity follows from the foundation axiom. We also have that it's chains are also well ordered because of the foundation axiom. By the theorem for each $b\in A$ it will be a subset of its union therefore we can find a minimal element allowing us to choose an element for each b. Is this true and is the proof correct?
Every partial order with well ordered chains is well founded is equivalent to the axiom of choice
axiom-of-choiceelementary-set-theoryorder-theorywell-orders
Related Solutions
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.
Of course we don't need the axiom of choice. In fact, there is nothing to prove.
Since you are considering $\bf E$ to be a class of ordered pairs, you are implicitly requiring that both $x$ and $y$ are sets, i.e. members of $\bf U$, otherwise they cannot be elements of an ordered pair.
Moreover, since $\overleftarrow{\bf E}(x)=x$ follows immediately from the Axiom of Extensionality, the Axiom of Foundation just tells you that $\bf E$ is well-founded.
To prove that $\in$ is also well-founded on classes, suppose that $X$ is a class, note that if $X$ is a set, then we have nothing to check, so we can assume it is a proper class, and so it is also non-empty.
Next, let $x_0\in X$ be some set, and now define $x_{n+1}=\bigcup x_n\cap X$, then $x_{n+1}$ is a set, by Union and Separation, and by Replacement, $\{x_n\mid n<\omega\}$ is also a set. Let $x=\bigcup\{x_n\mid n<\omega\}$, then it is a set, $x\subseteq X$, and so we have two options:
$x=\varnothing$, in which case we have that $x_0\subseteq x$, so $x_0=\varnothing$, so $x_0\cap X=\varnothing$.
Otherwise, by Foundation, there is some $y\in x$ such that $y\cap x=\varnothing$. However, if $y\in x$, then $y\in X$, and moreover, since $y\in x$ means $y\in x_n$ for some $n$, if $z\in y\cap X$, then $z\in x_{n+1}$. Therefore, since $y\cap x=\varnothing$, it must be that $y\cap X=\varnothing$ as well.
In either case, we found an element of $X$ which is disjoint from it.
Alternatively, you can just prove that assuming $\sf NBG$, we can write $\bf U$ as the union of its von Neumann hierarchy, $\{V_\alpha\mid\alpha\in\rm Ord\}$, so if $X$ is a proper class, there is a least $\alpha$ such that $X\cap V_\alpha$ is non-empty, and by minimality, any member of that intersection must be disjoint from $X$.
Best Answer
Every partial order has well-ordered chains, for example, singletons, or the empty chain.
Presumably you're asking about maximal well-ordered chains. And the answer to that is indeed positive.
It is easy to prove Zorn's lemma from this assumption: Assume that every partial order has maximal well-ordered chains, and let $(P,\leq)$ be a partial order in which every chain has an upper bound. If $C$ is a maximal well-ordered chain, let $x$ be an upper bound for it; if $x\notin C$, show that $C\cup\{x\}$ is still a well-ordered chain, so this is impossible, and therefore $x\in C$. From this follows that $x$ must be maximal, otherwise there would be an upper bound of $C$ which is not in $C$, which we established is impossible.
If what you're suggesting is "Every partial order whose chains are all well-ordered is well-founded", this is not enough to prove the axiom of choice.
To see that, first we need to prove another auxiliary choice principle: Dependent Choice, which is weaker than the axiom of choice. Dependent Choice can be stated as the principle: If $(P,\leq)$ is a partial order such that there is no infinite decreasing sequence, then $P$ is well-founded.
Now, assume Dependent Choice holds. If a partial order is not well-founded, then it has a decreasing sequence, which is a chain that is by definition not well-ordered, and therefore not all chains are well-ordered.
In the other direction, suppose that every partial order whose chains are well-ordered is well-founded, and let $(P,\leq)$ be a partial order without any decreasing sequences. Then consider $(Q,\prec)$ where $Q$ is the set of all decreasing sequences in $P$, ordered by reverse end-extensions (i.e., $\vec b\prec\vec a$ if $\vec a$ is an initial segment of $\vec b$).
It is not hard to verify that every chain in $Q$ is well-ordered, and therefore $Q$ is well-founded. Now suppose that $A$ is any subset of $P$, look at all the decreasing sequences in $A$, this is a subset of $Q$, so it has a minimal element, which is a decreasing sequence in $A$ that cannot be extended any further inside $A$, but those are finite sequences, and therefore this minimal sequence has an element which is minimal in $A$.