Question regarding a proof that: If $X, Y$ are ordinals, and $X\neq Y$, one is an Initial Segment of the other The proof states that if $X\subset Y$ or $Y\subset X$ then done.
So assume otherwise: then $X\cap Y\subset X$ and $X\cap Y\subset Y$. 
It then establishes a contradiction, dispelling this assumption to complete the proof.
My question is even if $X\subset Y$ or $Y\subset X$, wouldn't $X\cap Y\subset X$ and $X\cap Y\subset Y$ still be true. So how does it's negation establish the proof?
Thanks
EDIT As Prof. Caicedo suggests, here is the balance of the proof:
$X\cap Y$ is an ordinal (proved earlier) so $X\cap Y=X_a$ (an initial segment) and $X\cap Y=Y_b$. Then $a=X_a=X\cap Y=Y_b=b$. 
But $a\in X,b\in Y$. Thus $a=b\in X\cap Y$. But $X\cap Y=X_a$, so 
$x\in X\cap Y\rightarrow x\lt a$. In particular, $a\lt a$, a contradiction.
 A: Let's see. We already know that if $A,B$ are ordinals then so is $A\cap B$. Also, an initial segment of $A$ is either $A$ itself or else it has the form $A_a:=\{ x\in A:x<a\}$ for some $a\in A$. We also already have that, in that case,  $A_a=a$. And we also have that if $A\subseteq B$ then $A$ is an initial segment of $B$.
The problem is to check that if $X,Y$ are ordinals and $X\ne Y$ then one of them is a proper initial segment of the other. 
The proof is by contradiction. Hence, assume $X\nsubseteq Y$ and $Y\nsubseteq X$. Consider $X\cap Y$ and note that $X\cap Y\ne X$: otherwise, $X=X\cap Y\subseteq Y$ and therefore $X$ is after all an initial segment of $Y$. Since $X\cap Y$ is an ordinal and $X\cap Y\subseteq X$, then it is an initial segment of $X$ (and in fact a proper initial segment, since $X\cap Y\ne X$). Thus there is an $x\in X$ with $x=X_x=X\cap Y$.
The same argument, reversing the roles of $X$ and $Y$, shows that $X\cap Y$ is a proper initial segment of $Y$, so there is a $y\in Y$ with $y=Y_y=X\cap Y$.
Now we have reached a contradiction: We have $x=X\cap Y=y$. This means that $x=y$ is in both $X$ and $Y$ and therefore it is in $X\cap Y$. But no ordinal belongs to itself.
