I've attempted a proof of this which I believe to be somewhat unique, as I have not seen something similar here or else where.
X is an order topology.
Given a point $x_0$ $\in$ $X$ and a closed set $A$ such that ($x_0 \lt a $) $ \forall a \in A$, $\exists$ $N(x_0)$ such that $N(x_0) \cap A = \emptyset$
Define $B=X-N(x_0)$; $A \subseteq B$ as $\forall a \in A$, $a \not \in N(x_0)$.
Consider the case where $\exists a_0 \in A $ such that $(x_0,a_0) = \emptyset$
Construct $W = \cup(n,a_0)$ where $n \in N(x_0)$ and $V = \cup(x_0,b)$ where $b \in B$ and $x_0 \lt b$
By construction, $x_0 \in W$ and $A \in V$, where $V$ and $W$ are both open and $W \cap V = \emptyset$
Now, consider the case where $\exists d \in X$ such that ($x_0 \lt d \lt a$) $\forall a \in A$
$W = \cup(n,d)$ where $n \in N(x_0)$ and $V = \cup(d,b)$ where $b \in B$ and $x_0 \lt b$
By construction, $x_0 \in W$ and $A \in V$, where $V$ and $W$ are both open and $W \cap V = \emptyset$
In the case where $\exists a_1 \in A$ such that ($x \le a_1$) $\forall x \in X$ then $V = V \cup$ { $a_1$ } ; by construction, $V$ is now both closed and open, but this still satisfies the definition of regularity.
Similarly, if ($x_0 \le x$) $\forall x \in X$, then $W = W \cup$ { $x_0$ }, which is again a set that is both closed and open.
Repeat the construction of $V$ and $W$ in the case where ($a \lt x_0$) $\forall a \in A$ using the above method to complete the proof.
Does this proof work to prove regularity of an arbitrary order topology? the case where $x_0$ is the least element in the topology seems a bit sketchy in construction, and I don't know if the way I've done it is the most straight forward.
Thank you!
