Every Order Topology is Hausdorff I have the following proof:

Let $X$ be an ordered set having at least two elements with the order
  topology, and let $x,y \in X$. Without loss of generality, let $x < y$. If there is a $z \in X$ with
  $x < z < y$, then take $A = \{t \in X \mid t < z\}$ and $B = \{t \in X \mid t > z\}$. If not,
  take $A = \{t \in X \mid t < y\}$ and $B = \{t \in X \mid t > x\}$. Either way, note $x \in A$, 
  $y \in B$, and $A \cap B = \emptyset$, as needed.

Why in the second case is the intersection empty? I don't understand why this is true.

Suppose $z \in A \cap B$. Then $z < y$ (from $z \in A$) and $x < z$ (from $z \in B$). So then $x < z < y$ but no such point existed by assumption, as we are in the second case.  Contradiction, so $A \cap B = \emptyset$.
