Hausdorff's Maximal Principle implies Kuratowski's Lemma/Proof 1

Proof
Recall Hausdorff's Maximal Principle:

Recall Kuratowski's Lemma:

So, let us assume Hausdorff's Maximal Principle.

Let $S$ be a non-empty set of sets which is closed under chain unions.

$\set b$ is trivially a chain.

Hence by Hausdorff's Maximal Principle $\set b$ is a subset of a maximal chain of elements of $S$.

Hence $b$ is an element of a maximal chain $C$ of elements of $S$.

We have that $S$ is closed under chain unions.

Hence $\ds \bigcup C \in S$.

Then $b \subseteq \ds \bigcup C$.

If $\ds \bigcup C$ were a proper subset of some element $x \in S$, $C$ would be a proper subset of the chain $C \cup \set x$.

But $C$ is maximal so that cannot happen.

Hence $\ds \bigcup C$ is maximal.

Thus it is seen that Kuratowski's Lemma holds.