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.

Since there is no $z$ satisfying $x < z < y$, it follows that $y$ is the least upper bound for $x$ in this order. They are "next to" each other in the order in the way that $1$ and $2$ are "next to" each other in the usual order on the integers.
The set $A$ consists of all $t$ that are strictly less than $y$. The largest of these is $x$, so I think it would be more clear to write $A = \{t \in X \mid t \leq x\}$.
Similarly, the set $B$ consists of all $t$ that are strictly greater than $x$. The smallest of these is $y$, so I think it would be more clear to write $B = \{t \in X \mid t \geq y\}$.
The sets $A$ and $B$ can have no elements in common. Any element of $A$ is less or equal $x$, and therefore strictly less than $y$.
