Axiom of Choice implies Hausdorff's Maximal Principle/Proof 1

Proof
Let $S$ be the set of all chains of $\PP$.

$S \ne \O$ since the empty set is an element of $S$.

From Subset Relation is Ordering, we have that $\struct {S, \subseteq}$ is partially ordered by inclusion.

Let $C$ be a totally ordered subset of $\struct {S, \subseteq}$.

Then $\bigcup C$ is a chain in $C$ by Set of Chains is Chain Complete for Inclusion.

This shows that $S$, ordered by set inclusion, is an inductive ordered subset.

By applying Zorn's Lemma, the result follows.