(Verification) Zorn's Lemma is Equivalent to Hausdorff Maximal Principle 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 $X$ 
 A: It could be better to isolate the main parts of the proof in two lemmas that show a bit more than what you have to actually prove.
Definition. If $(X,\le)$ is a partially ordered set, denote by  $\hat{X}$ the set of chains in $X$, ordered by inclusion.
In what follows, $(X,\le)$ is supposed to be a poset. The next lemma uses the fact that the union of a chain of chains is again a chain and is interesting on its own, because it shows $\hat{X}$ always satisfies a strong version of Zorn’s lemma hypotheses.
Lemma 1. Any chain in $\hat{X}$ has a least upper bound.
Proof. Suppose, for $i\in I$, that $C_i$ is a chain in $X$ and that, for $i,j\in I$, we have $C_i\subseteq C_j$ or $C_j\subseteq C_i$. Let $\varphi(i,j)=j$ if $C_i\subseteq C_j$ and $\varphi(i,j)=i$ if $C_j\subseteq C_i$.
Consider $C=\bigcup_{i\in I}C_i$. If $x,y\in C$, then $x\in C_i$ and $y\in C_j$, for some $i,j\in I$. Then $x,y\in C_{\varphi(i,j)}$, so either $x\le y$ or $y\le x$. Hence $C$ is a chain in $X$ and is obviously the least upper bound of the given chain. QED
Lemma 2. If $C\in\hat{X}$ is maximal, then it has at most one upper bound, which is maximal in $X$.
Proof. If $u$ is an upper bound of $C$, then the set $C\cup\{u\}$ is a chain and contains $C$; by maximality, $u\in C$, so $u=\max C$. Non maximality of $u$ would contradict $u$ being the maximum of $C$. QED
Now note that Lemma 1 proves that Zorn’s lemma implies Hausdorff’s maximal principle. The converse is proved by Lemma 2.
A: 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.
