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
As Noah Schweber pointed out, if $X$ is a partially ordered set that satisfies the hypothesis of Zorn's lemma, then $X$ is not empty:
If $X$ is the empty set, then there is precisely one chain $A$ - the empty set - and this chain does not have an upper bound in $X$, since $X$ is empty.$^1$ But the hypothesis is that each chain has an upper bound, so we get a contradiction.
$^1$ However, if $X$ is nonempty, then the empty chain does have an upper bound: "If $A$ is a chain in $X$, the hypothesis of Zorn's lemma guarantees the existence of an upper bound for $A$ in $X$; it does not guarantee the existence of an upper bound for $A$ in $A$" (see the discussion below the statement of Zorn's lemma on page 62).