Axiom of Choice implies Hausdorff's Maximal Principle/Proof 2

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

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

Proof
Let $\preceq$ be an ordering on the set $\mathcal P$.

Let $X$ be a chain in $\left({\mathcal P, \preceq}\right)$.

By definition, a maximal chain in $\mathcal P$ that includes $X$ is a chain $Y$ in $\mathcal P$ such that $X \subseteq Y$ and there is no chain $Z$ in $\mathcal P$ with $X \subseteq Z$ and $Y \subsetneq Z$.

Let us define $\mathcal C$ as:
 * $\mathcal C = \left\{{Y : Y}\right.$ is a chain in $\mathcal P$ and $\left.{X \subseteq Y}\right\}$

Then: are one and the same.
 * a maximal chain in $\mathcal P$ that includes $X$ and
 * a maximal element of $\mathcal C$ under the partial order induced on $\mathcal C$ by inclusion

It thus suffices to show that $\mathcal C$ contains such a maximal element.

According to Zorn's Lemma, $\mathcal C$ contains a maximal element if each chain in $\mathcal C$ has an upper bound in $\mathcal C$.

Let $W$ be a chain in $\mathcal C$, and let $Z = \bigcup W$.

Let $a, b \in Z$.

Then $\exists A, B \in W: a \in A, b \in B$.

Since $W$ is a chain in $\mathcal C$, one of $A$ and $B$ includes the other.

Suppose WLOG $A \subseteq B$.

Then $a, b \in B$.

Since $B$ is a chain in $\mathcal P$, either $a \preceq b$ or $b \preceq a$.

Therefore $Z$ is a chain in $\mathcal P$.

By definition of $Z$, we have that $\forall A \in W: A \subseteq Z$.

Thus $Z$ is an upper bound for $W$ under inclusion.

Finally we have that:
 * $\forall A \in W: X \subseteq A$

and so $X \subseteq Z$.

Thus $Z \in \mathcal C$