Question about Hausdorff Maximal principle and antichain I have this prob and still haven't figured about

Let (A, $\leq$) be a partially ordered set. A subset B of A is called an antichain if and only if for any two distinct elements x and y in B, neither x $\leq$ y or y $\leq$ x. Prove that every antichain is contained in an antichain which is maximal with respect to inclusion $\subseteq$

I tried to make a set of subsets of A which are antichains. However I don't know what to do next. The text book says use Hausdorff maximal principle, but I can't use it from the set I made cause elements are not totally ordered sets. What should I do?
 A: Let $L$ be the set of all antichains of $A$. $L$ is evidently non-empty, since every singleton subset $\{x\}\subset A$ is an antichain. 
Since $L \subset \mathcal{P}A$ we have that $L$ is partially ordered by set inclusion. 
For an antichain $B$, $\{B\}\subseteq L$ is totally ordered by set inclusion trivially. Hence applying Hausdorff Maximal principle it is contained in a maximal totally order chain which we call $\mathcal{C}\subseteq L$. 
Let $\bar{B} = \cup \mathcal{C}$. 
Verify $\bar{B}$ is maximal: if not, there would be $C$ an antichain such that $C\supsetneq \bar{B}$. But then $\mathcal{C} \cup \{C\} \supsetneq \mathcal{C}$ contradicts maximality of $\mathcal{C}$. 
Next we verify $\bar{B}$ is an antichain. If $x,y\in \bar{B}$ we know that $\exists B_x,B_y\in \mathcal{C}$ such that $x\in B_x, y\in B_y$. But since $\mathcal{C}$ is totally ordered we have that $\exists B' \in \mathcal{C}$ such that $B' \supset B_x \cup B_y$ and so $x,y\in B'$. Since $B'\in L$ we have that $x,y$ are not comparable. Hence $\bar{B}$ is an antichain. 
