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

Theorem
Let $\left({\mathcal P, \preceq}\right)$ be a partially ordered set.

Then there exists a maximal chain in $\mathcal P$.

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

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

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

Let $C$ be a totally ordered subset of $\left({S, \subseteq}\right)$.

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 poset.

By applying Zorn's Lemma, the result follows.