Let $(X, \le)$ be a partially ordered set $X$.
Claim
Zorn's Lemma and Hausdorff Maximal Principle are equivalent.
Zorn's Lemma
Suppose $X$ has the property that every chain has an upper bound in $X$. Then the set $X$ contains at least one maximal element.
Hausdorff Maximal Principle
$X$ holds maximal chain.
$1.\;$Zorn's Lemma $\rightarrow$ Hausdorff Maximal Princple
Let $\Bbb C(X)$ be the family of every chain of $X$ and Let $\Bbb C$ be the chain of $\Bbb C(X)$ and Let $C= \cup\Bbb C$.
Then for $a,b, \in C$ there $\exists C_1, C_2 \;\text{s.t}. a \in C_1 \in \Bbb C \;\text{and}\; b \in C_2 \in \Bbb C$
But $C_1 \subset C_2 \;\text{or}\;C_2\subset C_1 $ since $C$ is chain.
If $C_1 \subset C_2 $, $a,b \in C_2$. Then $a \le b \;\text{or}\; b \le a$ since $C_2$ is a chain of $X$ and $vice\;versa$
Thus $C$ is a chain of $X$
Now Hausdorff Maximal Principle holds since $\Bbb C \subset \Bbb C(X)$ has maximal chain $C$
$2.\;$ Hausdorff Maximal Princple $\rightarrow$ Zorn's Lemma
Suppose every chain of $X$ has an upper bound. Then for the maximal chain of $X$,$\;C$, let $m\in X$ be the upperbound of $C$.
Now suppose $x \in X \;\text{and}\; x>m$ Then
$C \cup \{x\}$ is also a chain since x is comparable with an element in $C$
But it contradicts to the fact that $C$ is maximal chain since $C \cup \{x\} \supsetneq C$
Thus m is a maximal element of $X$
Best Answer
For 1. I would write: Let $C(X)$ be the set of chains of $X,$ partially ordered by $\subset$. Then show that $\subset$ is a transitive relation on $C(X).$ (Which is fairly obvious)...Then show, as you did that if $S$ is a $\subset$-chain of $C(X),$ then $\cup S\in C(X)$ and $\cup S$ is a $\subset$-upper bound for $S$. Zorn's Lemma then implies that $C(X)$ has a $\subset$-maximal member....Part 2 is OK.
As I said in a comment, it's my opinion that your presentation could be a bit better.