A proof of Zermelo’s Well-Ordering Theorem using Zorn’s Lemma The thing is that I have used Zorn’s Lemma to prove that, given a set $A$, it can be well ordered, but I don’t know if I’ve made any wrong assumptions.
Let $A$ be a set, and we want to prove that it exists a well-order relation for $A$. Consider the following set: $Z = \{B\subseteq A | B \textrm{ can be well ordered}\}$. The partial order of $Z$ is the following: since every well ordered set is isomorphic to a single ordinal, given $B \textrm{ and }C\in Z$, we will say that $B\leq _Z C$ if $B$ is isomorphic to an ordinal $\alpha$, $C$ is isomorphic to an ordinal $\beta$ and $\alpha \leq \beta$. Since $A\neq \emptyset$, it exists some $B\subseteq A$ such that $|B|<\aleph _0$. In particular, $B$ can be well-ordered, so $B\in Z$ and $Z\neq \emptyset$.
Now we are going to prove that every chain of $Z$ has an upper bound in $Z$. Let $C$ be a chain. It is enough to prove that $\cup C \in Z$ because it is obvious that in that case $\cup C$ would be an upper bound of $C$.

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*$\cup C \subseteq A$: if $x \in \cup C$, it exists $B \in C$ such that $x\in B \subseteq A$, so $x \in A$.

*$\cup C$ can be well ordered: (here is where I have some doubts) but I think it is immediate that: $$\cup C \approx sup \{\alpha |\exists B \in C \textrm{ such that }\alpha \approx B\}$$ where “$\approx”$ means that the sets are equipotent. Since the supremum of a set of ordinals is an ordinal, we see that $\cup C$ is equipotent to an ordinal, let say $\gamma$. In particular, the bijection between $\cup C$ and $\gamma$ gives us a well order for $\cup C$. We conclude that $\cup C \in Z$.

We have proved that every chain of $Z$ has an upper bound in $Z$ so appealing to Zorn’s Lemma, $Z$ has a $\leq _Z$-maximal element, let say $M$. The only thing we have to prove now is that $M = A$. If not, $A\setminus M \neq \emptyset$, so let $x\in A\setminus M$. If $M$ is isomorphic to an ordinal $\alpha$ then, since $x\notin M$, $M\cup \{x\}$ is isomorphic to $\alpha ^+$. This is not possible because $\alpha <\alpha ^+$, contradicting the maximality of $M$. So $M = A$ and we have proven that the set $A$ can be well ordered.
Any suggestion is welcome!
 A: Your proof is at least kind of circular. If a set is infinite and can be well-ordered, then it can be well-ordered in a lot of different ways, and even for finite sets the particular choice of order requires you to choose the order.
Suppose that $X$ is an infinite Dedekind-finite set which is a countable union of pairs, i.e. a Russell/socks set. If $X_n$ is the $n$th pair, then $A_n=\bigcup_{k<n}X_k$ will be a chain of finite sets without an upper bound, since the union is $X$ itself, which cannot be well-ordered.
The point is that $\sf ZF$ cannot prove that the union of sets which can be well-ordered is itself a set that can be well-ordered.1
The correct approach is to consider subsets and well-orders of the subset, and have the extension as end-extensions. Or, slightly more straightforward, simply consider injective functions from an ordinal into the set.


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*Interestingly, in Cohen's first model, where the axiom of countable choice fails, it is true that the union of well-orderable sets is well-orderable. So as far as axioms go, this one is pretty weak. On the other hand, Dependent Choice, which is much stronger than countable choice, is not enough to prove that the union of $\aleph_1$ sets of size $\aleph_1$ (or indeed, pairs!) can be well-ordered.
So really, this choice principle is somehow "orthogonal" to classical choice axioms.
