Let $X$ be a set partially ordered by $<$ such that each chain has an upper bound. Define $p(x)=\{y\in X\mid x< y\}\in P(X)$ . Let $p(X)=\{p(x)\mid x\in X\}$ . If $p(x)=\emptyset$ then it follows that $x$ is maximal.

Suppose no $p(x)=\emptyset$ . Then by the axiom of choice there is a choice function $f$ on $p(X)$ , and since for each $p(x)$ we have $f(p(x))\in p(x)$ , it follows that $x<f(p(x))$ . Define $f_\alpha(p(x))$ for all ordinals $\alpha$ by transfinite induction:

$f_0(p(x))=x$

$f_{\alpha+1}(p(x))=f(p(f_\alpha(p(x))))$

And for a limit ordinal $\alpha$ , let $f_\alpha(p(x))$ be an upper bound of $f_i(p(x))$ for $i<\alpha$ .

This construction can go on forever, for any ordinal. Then we can easily construct an injective function from $Ord$ to $X$ by $g(\alpha)=f_\alpha(p(x))$ for an arbitrary $x\in X$ . This must be injective, since $\alpha<\beta$ implies $g(\alpha)<g(\beta)$ . But that requires that $X$ be a proper class, in contradiction to the fact that it is a set. So there can be no such choice function, and there must be a maximal element of $X$ .

For the reverse, assume Zorn's lemma and let $C$ be any set of non-empty sets. Consider the set of functions $F=\{f\mid \forall a\in\operatorname{dom}(f) (a\in C \wedge f(a)\in a)\}$ partially ordered by inclusion. Then the union of any chain in $F$ is also a member of $F$ (since the union of a chain of functions is always a function). By Zorn's lemma, $F$ has a maximal element $f$ , and since any function with domain smaller than $C$ can be easily expanded, $\operatorname{dom}(f)=C$ , and so $f$ is a choice function for $C$ .