Order Topology is Hausdorff

Theorem
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.

Then $\struct {X, \tau}$ is a Hausdorff space.

Proof
Let $x, y \in X$ with $x \ne y$.

Since $\le$ is a total ordering, either $x \prec y$ or $y \prec x$.

, assume that $x \prec y$.

If there is a $z \in X$ such that $x \prec z \prec y$, then $z^\prec$ and $z^\succ$ separate $x$ and $y$.

Otherwise, by Upper Closure is Strict Upper Closure of Immediate Predecessor, $y^\prec$ and $x^\succ$ separate $x$ and $y$.

Since any two distinct points can be separated by neighborhoods, $\struct {X, \tau}$ is a Hausdorff space.