I'm trying to prove the below which is a problem in my text book.
If $\mathcal{A}= \langle A,\le\rangle$ is a partially ordered set then $\mathcal{A}$ has a maximal chain $\Leftrightarrow$ If $\mathcal{A}=\langle A,\le\rangle$ is a partially ordered set and $B$ is a chain of $\mathcal{A}$ then there is a $C$ such that $B\subseteq C$ and $C$ is a maximal chain of $\mathcal{A}$. (This statement is to show the equivalence of a statement to the Hausdorff maximal principle (so we cannot assume the AC)).
I'm new to these maximal principles. Here is my attempted proof. I'd be grateful if anyone could let me know if I am on the right track:
$(\Leftarrow)$ Suppose $a\in A$. Then $ \{a\} $ is a chain of $\mathcal{A}$ and so there is a $C$ such that $\{a\} \subseteq C$ and $C$ is a maximal chain of $\mathcal{A}$. Hence $\mathcal{A}$ has a maximal chain.
$(\Rightarrow)$ Suppose $B$ is a chain of $\mathcal{A}$. Let $\mathcal{E}=\{D |B\subseteq D$ and$\ D$ is a chain of $\mathcal{A}\}$ . Then, $\bigcup\mathcal{E}\subseteq A$ and so $\langle\bigcup\mathcal{E}, \le\rangle$ is a partially ordered set. Hence by our assumption, there is a maximal chain of $\langle\bigcup\mathcal{E},\le\rangle$, say $C$. We claim $B\subseteq C$. Suppose $b\in B$. Then as $B\in\mathcal{E}, b\in\bigcup\mathcal{E}$. Suppose $b\notin C$. Then we claim $C\cup\{b\}$ is a chain of $\bigcup\mathcal{E}$. We know $C\cup\{b\}\subseteq\bigcup\mathcal{E}$ and $C\subset C\cup\{b\}$. And if $x,y\in C\cup\{b\}$ then either a) $x,y\in C$ in which case $x$ and $y$ are comparable as $C$ is a chain, b) $x\in C,y=b$ in which case $x\in\bigcup\mathcal{E}$ so $x\in D$ for some $D\in\mathcal{E}$. And as $B\subseteq D$, $b\in D$ and as $D$ is a chain, $x$ and $y$ are comparable c) $x=b,y\in C$, in which case $x$ and $y$ are comparable by the same reasoning or d) $x=b,y=b$ in which case $x$ and $y$ are comparable as $\le$ is reflexive in $A$.Hence as $C\cup\{b\}\subseteq\bigcup\mathcal{E}$ and $C\cup\{b\}$ is connected, $C\cup\{b\}$ is a chain of $\langle \bigcup\mathcal{E}, \le \rangle$, a contradiction as $C$ is a maximal chain of $\langle \bigcup\mathcal{E}, \le \rangle$. Hence$\ b\in C$ and $B\subseteq C$.
We will now show $C$ is a maximal chain of $\mathcal{A}$. Suppose $C\subset E$ and $E$ is a chain of $\mathcal{A}$. As $B\subseteq C$, $B\subseteq E$ and as $E$ is a chain of $\mathcal{A},E\in\mathcal{E}$, and so $E \subseteq \bigcup\mathcal{E}$. Hence $E$ is a chain of $ \langle \bigcup\mathcal{E},\le \rangle$, a contradiction to maximality of $C$. Hence $C$ is a maximal chain of $\mathcal{A}$.
This proof parallels Exercise 5.4.4 in a sense.
$(\Leftarrow)$Let $B = \varnothing$, then the proposition is exactly the Hausdorff's Maximal Principle(theorem 5.14). $(\Rightarrow)$Let $\mathscr{S}_{B} = \{C \in \mathscr{S} \mid B \subseteq C\}$ where $\mathscr{S}$ is the set of all chains of $A$, ordered by inclusion. By hypothesis, $\mathscr{S}_{B}$ has a maximal chain, it is also a maximal chain of $\mathscr{S}$.