Axiom of Choice implies Zorn's Lemma/Proof 2

Theorem
Acceptance of the Axiom of Choice implies the truth of Zorn's Lemma.

Statement of Zorn's Lemma
Let $\left({X, \preceq}\right), X \ne \varnothing$ be a non-empty ordered set such that every non-empty chain in $X$ has an upper bound in $X$.

Then $X$ has at least one maximal element.

Proof
Suppose for the sake of contradiction 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 for each $x \in X$, $x \prec f(x)$.

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

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

Thus by the Axiom of Choice, there is a mapping $g: \mathcal C \to X$ such that for each $C \in \mathcal C$, $g(C)$ 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 $\operatorname{On} \times h(\operatorname{On})$.

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

By the Axiom of Replacement, $\operatorname{On}$ 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.