Axiom of Choice implies Zorn's Lemma/Proof 2

Proof
that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.

By the Axiom of Choice, there is a mapping $f: X \to X$ such that:
 * $\forall x \in X: x \prec \map f x$

Let $\CC$ be the set of all chains in $X$.

By the premise, each element of $\CC$ has an upper bound in $X$.

Thus by the Axiom of Choice, there is a mapping $g: \CC \to X$ such that for each $C \in \CC$, $\map g C$ is an upper bound of $C$.

Let $p$ be an arbitrary element of $X$.

Define a mapping $h: \operatorname {Ord} \to X$ by transfinite recursion thus:

Then $h$ is strictly increasing, and thus injective.

Let $h'$ be the restriction of $h$ to $\On \times \map h \On$.

Then ${h'}^{-1}$ is a surjection from $\map h \On \subseteq X$ onto $\On$.

By the Axiom of Replacement, $\On$ is a set.

By Burali-Forti Paradox, this is a contradiction.

Thus we conclude that some element of $X$ has no strict successor, and is thus maximal.