Hausdorff Maximality Principle

Theorem
Let $$X$$ be a Partially Ordered Set.

Then there exists a maximal totally ordered subset (Maximal Chain) of $$X$$.

Proof
Let $$S$$ be the set of all chains of $$X$$.

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

From Subset Relation on Set of Subsets is an 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.