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 partial ordered by inclusion.

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

Since each set in $c$ is itself a chain, and hence an element of $S$, the union of these chains is also a chain, and an element of $S$.

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

By applying Zorn's Lemma, the result follows.