Axiom of Choice implies Kuratowski's Lemma/Proof 1

Proof
Let $S$ be a (non-empty) ordered set.

Let $C$ be a chain in $S$.

Let $P$ be the set of all chains that are supersets of $C$.

Let $\CC$ be a chain in $\powerset P$ (partially ordered by set-inclusion).

Define $C' = \bigcup \CC$.

Note that the elements of $P$ are chains on $\paren S$, so the elements of $\CC$ are also chains in $S$, as $\CC$ is a subset of $P$.

Thus $\bigcup \CC$ contains elements in $S$, so:
 * $C' \subseteq S$.

First, note that $C'$ is a chain in $S$.

Let $x, y \in C'$, which means $x \in X$ and $y \in Y$ for some $X, Y \in \CC$.

However, as $\CC$ is a chain in $\powerset P$, that means either $X \subseteq Y$ or $Y \subseteq X$.

So $x$ and $y$ belong to the same chain in $S$.

Thus either $x \le y$ or $y \le x$.

Thus $C'$ is a chain on $S$.

Now let $x \in C$.

Then:
 * $\forall A \in P: x \in A$

Then because $\CC \subseteq P$:
 * $\forall A \in \CC: x \in A$

So:
 * $x \in \bigcup \CC$

and so $C \subseteq C'$

Thus:
 * $C' \in P$

Now, note $C'$ is an upper bound on $\CC$.

To prove this consider $x \in D \in \CC$.

This means:
 * $x \in \bigcup \CC = C'$

so:
 * $D\subseteq C'$

The chain in $P$ was arbitrary, so every chain in $P$ has an upper bound.

Thus, by Zorn's Lemma, $P$ has a maximal element.

This must be a maximal chain containing $C$.