Proof of the Hausdorff maximal principle from Zorn's lemma

Proof

We apply Zorn's lemma to a set of chains ordered by inclusion.

1. Let $(P, \le)$ be a partially ordered set and let $C_0$ be a chain of $P$.
2. Let $Q$ be the set of chains $C$ of $P$ with $C_0 \subseteq C$. Order $Q$ by inclusion $\subseteq$, which is a partial order. The set $Q$ is nonempty, since $C_0 \in Q$.
3. Let $K$ be a chain of $(Q, \subseteq)$. If $K$ is empty, then $C_0$ is an upper bound of $K$ in $Q$.
4. Let $K$ be nonempty, and let $W = \bigcup_{C \in K} C$ be the union of its members. Then $C_0 \subseteq W$.
5. The set $W$ is a chain of $P$. Let $x, y \in W$, with $x \in C$ and $y \in D$ for some $C, D \in K$. Since $K$ is a chain, $C \subseteq D$ or $D \subseteq C$.
6. So $x$ and $y$ lie in the larger of $C$ and $D$, which is a chain of $P$. Hence $x \le y$ or $y \le x$.
7. So $W \in Q$, and $C \subseteq W$ for each $C \in K$. Thus $W$ is an upper bound of $K$ in $Q$.
8. By steps 2 to 7, Zorn's lemma applies to $(Q, \subseteq)$. It gives a maximal element $M$ of $Q$.
9. Then $M$ is a chain with $C_0 \subseteq M$. Suppose that $D$ is a chain of $P$ with $M \subsetneq D$.
10. Then $C_0 \subseteq D$, so $D \in Q$, and $M \subsetneq D$ contradicts the maximality of $M$ in $Q$.
11. So $M$ is a maximal chain of $P$ that contains $C_0$.