Greets
This is a problem I wanted to solve for a long time, and finally did some days ago. So I want to ask people here at MSE to show as many different answers to this problem as possible. I will offer a Bounty in two days, depending on the interest in the problem, and eventually increase it as it gets more voted. Of course, I will show my answer to this question to know whether it is correct.
Thanks
$\newcommand{\cl}{\operatorname{cl}}$ Theorem. Let $\langle X,\tau,\le\rangle$ be a LOTS; then $X$ is $T_5$.
Proof. Let $H$ and $K$ be separated subsets of $X$. For each $x\in H$ there is a convex $V_x\in\tau$ such that $x\in V_x\subseteq X\setminus K$, and for each $x\in K$ there is a convex $V_x\in\tau$ such that $x\in V_x\subseteq X\setminus H$. Let $V_H=\bigcup_{x\in H}V_x$ and $V_K=\bigcup_{x\in K}V_x$; clearly $H\subseteq V_H$, $K\subseteq V_K$, and $V_H\cap V_K\subseteq X\setminus(H\cup K)$.
Let $V=V_H\cap V_K$. If $V=\varnothing$, we’re done, so suppose that $V\ne\varnothing$. Define a relation $\sim$ on $V$ by $p\sim q$ iff $\left[\min\{p,q\},\max\{p,q\}\right]\subseteq V$; it’s easily verified that $\sim$ is an equivalence relation whose equivalence classes are the order-components of $V$.
Let $T\subseteq V$ contain exactly one point of each $\sim$-class. Suppose that $x\in H$, and $p,q\in V_x\cap T$ with $p<q$; I’ll show that $p<x<q$. Suppose that $x<p$. Since $p\in T\subseteq V$, there is a $y\in K$ such that $p\in V_y$; $[x,q]\subseteq V_x\subseteq X\setminus K$, so $y\notin[x,q]$. If $y<x$, then $x\in[y,p]\subseteq V_y\cap H=\varnothing$, so $x<p<q<y$. But then $[p,q]\subseteq V_x\cap V_y\subseteq V$, so $p\sim q$, contradicting the choice of $p$ and $q$ and showing that $p<x$. A similar argument shows that $x<q$. Similarly, if $x\in K$ and $p,q\in V_x\cap T$ with $p<q$, then $p<x<q$. Note that it follows immediately that $|V_x\cap T|\le 2$ for all $x\in H\cup K$.
Now fix $p\in T$. Let $H_p=\{x\in H:p\in V_x\}$ and $K_p=\{x\in K:p\in V_x\}$; $H_p\ne\varnothing\ne K_p$, since $p\in V$. Suppose that $x<p$ for some $x\in H_p$. If $y\in K_p$ and $y<p$, then either $x<y$ and $y\in V_x$, or $y<x$ and $x\in V_y$, since the sets $V_x$ and $V_y$ are convex; neither is possible, so $p<y$, and since $y\in K_p$ was arbitrary, $p<K_p$. A similar argument then shows that $H_p<p$ and hence $H_p<p<K_p$. If instead $p<x$ for some $x\in H_p$, it follows similarly that $K_p<p<H_p$.
For each $x\in H\cup K$ define $W_x\in\tau$ as follows: $$W_x=\begin{cases}V_x,&\text{if }V_x\cap T=\varnothing\\V_x\cap(p,\to),&\text{if }V_x\cap T=\{p\}\text{ and }p<x\\V_x\cap(\leftarrow,p),&\text{if }V_x\cap T=\{p\}\text{ and }x<p\\V_x\cap(p,q),&\text{if }V_x\cap T=\{p,q\}\text{ and }p<x<q\;.\end{cases}$$ Let $$W_H=\bigcup_{x\in H}W_x\qquad\text{and}\qquad W_K=\bigcup_{x\in K}W_x\;;$$ clearly $W_H$ and $W_K$ are open, $H\subseteq W_H$, and $K\subseteq W_K$, and I claim that $W_H\cap W_K=\varnothing$.
Suppose not; then there are $x\in H$ and $y\in K$ such that $W_x\cap W_y\ne\varnothing$; without loss of generality suppose that $x<y$. Fix $q\in W_x\cap W_y$; it’s not hard to see that $x<q<y$, since $W_x$ and $W_y$ are convex. Moreover, $q\in V$, so $q\sim p$ for a unique $p\in T$. Let $I$ be the closed interval with endpoints $p$ and $q$. Then $I\subseteq V\subseteq X\setminus(H\cup K)$, so $x,y\notin I$, and therefore $x<p<y$. If $p\le q$, then $p\in W_x\cap T\subseteq V_x\cap T$, and by construction $W_x\subseteq(\leftarrow,p)$, and $p\notin W_x$, a contradiction. If, on the other hand, $q\le p$, then $p\in W_y\cap T\subseteq V_y\cap T$, so that $W_y\subseteq(p,\to)$, and $p\notin W_y$, which is again a contradiction, and it follows that $W_H\cap W_K=\varnothing$. $\dashv$
Despite the sometimes finicky details, the idea of the argument is very simple. $V$ is the intersection of the open nbhds $V_H$ and $V_K$ of $H$ and $K$, respectively. It’s open, so we partition it into its open order-components. If $C$ is one of these components, let $H_C=\{x\in H:V_x\cap C\ne\varnothing\}$ and $K_C=\{x\in K:V_x\cap C\ne\varnothing\}$; we show that either $H_C<C<K_C$ or $K_C<C<H_C$. Then we pick a point $p$ in $C$ and contract the intervals $V_x$ that meet $C$ by intersecting them with $(\leftarrow,p)$ for $x<C$ and with $(p,\to)$ for $C<x$. This contraction removes $C$ from the intersection of the nbhds of $H$ and $K$, and since we do it simultaneously for all components $C$, we end up with disjoint nbhds $W_H$ and $W_K$ of $H$ and $K$.
As an immediate consequence we get that every LOTS $X$ is hereditarily normal: disjoint closed sets in a subspace of $X$ are separated sets in $X$. Thus, we have for free that every GO-space (generalized ordered space) is hereditarily normal.
Moreover, the argument is very easily modified to show that if $\mathscr{F}$ is a separated family of subsets of $X$, meaning that $F\cap\cl\bigcup(\mathscr{F}\setminus\{F\})=\varnothing$ for each $F\in\mathscr{F}$, then there is a pairwise disjoint family $\mathscr{U}=\{U_F:F\in\mathscr{F}\}$ of open sets in $X$ such that $F\subseteq U_F$ for each $F\in\mathscr{F}$, i.e., that $X$ is hereditarily collectionwise normal.
Proposition 1. If $\langle X,<\rangle$ is a connected linear order with endpoints, then the ordered space $X$ is compact.
Proof. Note that under these hypothesis every nonempty subset of $X$ has a supremum. Hence this property is proved using the idea in the classic proof that $[0,1]$ is compact; which is done considering an open covering $\Lambda$ of $[0,1]$ and showing that $\sup\{x\in [0,1]:[0,x]$ can be covered by a finite subset of $\Lambda\}$ equals $1$; using the completeness of $[0,1]$.
Proposition 2. If $\langle X,< \rangle$ is a linear order, then $\langle X,< \rangle$ can be embedded in a connected order without endpoints.
Proof. First embed $\langle X,< \rangle$ in a dense linear ordering $\langle X',< \rangle$ without endpoints; which is pretty easy, and then using Dedekind cuts, $\langle X',< \rangle$ can be embedded in a connected linear order without endpoitns $\langle X'',< \rangle.$
Proposition 3. There exists a family $M$ of closed intervals of $(X,<)$ such that the intersection of any two elements of $M$ is at most a point and $\bigcup M=X$.
Fix $x\in X$ not an endpoint. Put $x_0=x$, and let $x_0\in X$ be such that $x<x_0$. Suppose that for some ordinal $\alpha$ an increasing $\alpha$-sequence $\langle x_{\beta}:\beta < \alpha \rangle$ such that $x_{\gamma}<x_{\beta}$ whenever $\gamma<\beta$ has been constructed. Let $x_{\alpha}\in X$ be such that $x_{\alpha}$ is greater than all elements of $\langle x_{\beta}:\beta < \alpha \rangle$, if such $x$ exists. But $X$ is a set, so there exists some ordinal $\alpha$ such that the construction cannot be continued at step $\alpha$. Then we have $[x,\infty)=\bigcup_{\beta<\alpha}[x_{\beta},x_{\beta+1}]$, similarly there is a family of closed intervals $\mathfrak{I}$ such that the intersection of any two elements of $\mathfrak{I}$ is at most a point and such that $\bigcup \mathfrak{I}=(\infty,x],$ then put $M=\{[x_{\beta},x_{\beta+1}]:\beta<\alpha\}\cup \mathfrak{I}$.
Proposition 4. If $\langle X,<\rangle$ is connected, then $\langle X,< \rangle$ is normal.
Proof. Let $M$ be a family of closed intervals with the property of the previous proposition. Then each $I\in M$ is compact by Proposition 1, but also each $I\in M$ is Hausdorff, hence each $I\in M$ is a normal subspace of $\langle X,<\rangle$. Now let $A,B$ be closed disjoint subsets of $\langle X,<\rangle$. For each $I\in M$ let $A_I$ and $B_I$ be disjoint open subsets of $I$ with $A\cap I\subseteq A_I$ and $B\cap I\subseteq B_I$. Furthermore the sets $A_I,B_I$ can be chosen so that if $A$ does not contain an endpoint,$x$, of $I$, then $x\notin A_I$, and $B_I$ can also be chosen so that this holds, this ensures that for distinct $I,J\in M$, $A_I\cap B_J=\emptyset$; since $A\cap B=\emptyset$ and $I$ and $J$ have only at most one point in common. Hence if $A'=\bigcup_{I\in M}A_I$ and $B'=\bigcup_{I\in M}B_I$, $A'\cap B'=\emptyset$, and both $A'$ and $B'$ are open subsets of $X$ separating $A$ and $B$; since $\bigcup M=X$. Therefore $\langle X,<\rangle$ is normal.
Proposition 5. Every ordered space is normal.
Proof. Let $\langle X,<\rangle$ be a linear order, and let $\langle X',<\rangle$ be connected linear order without endpoints exteding $\langle X,<\rangle$. Let $A,B\subset X$ be disjoint and closed, then if $Y=X'-(\operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B))$, $A,B\subset Y$ since $X\cap \operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B)=\emptyset$, because of the hypothesis, and so $X\subseteq Y$. But $Y$ is open in $X'$, hence $Y$ is the union of a disjoint family of open intervals, each being connected, hence by Proposition 4, $Y$ is a normal subspace of $X$. But $\operatorname{Cl}_{Y}(A)\cap \operatorname{Cl}_{Y}(B)=\operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B)\cap Y=\emptyset$, thus there are disjoint open subsets $A',B'$ of $Y$ such that $\operatorname{Cl}_{Y}(A)\subseteq A'$ and $\operatorname{Cl}_{Y}(B)\subseteq B'$. Therefore $A'\cap X$ and $B'\cap X$ are disjoint open subsets of $X$ separating $A$ and $B$, respectively. Thus $\langle X,<\rangle$ is normal.
(Generalised) ordered spaces (GO-spaces) are monotonically normal, and monotonically normal spaces are hereditarily normal. I wrote proofs here.
Let $\langle X,\tau,\le\rangle$ be a LOTS.
If A and B are nonempty subsets of $X$, $A \lt B$ shall mean that $a \lt b$ for each $a \in A$ and $b \in B$. When this relation holds, every point in $B$ is an upper bound for the set $A$, and, every point in $A$ is a lower bound for the set $B$.
Theorem 1: For $A \lt B$ there exist open sets $R_A$ and $R_B$ such that
$\quad A \subset R_A$
$\quad B \subset R_B$
$\quad R_A \lt R_B$
IF AND ONLY IF
$\quad$If $A$ has no greatest element and $B$ has a least element $b$, then
$\quad$point $b$ can't be the supremum of the set $A$
$\quad$and
$\quad$If $B$ has no least element and $A$ has a greatest element $a$, then
$\quad$point $a$ can't be the infimum of the set $B$
Proof:
The necessity is straightforward. The sufficiency is also straightforward when you can 'use' the premises. But you also must construct $R_A$ and $R_B$ for the two other cases:
$A$ has a maximum AND $B$ has a minimum
$A$ has no maximum AND $B$ has no minimum
But each of these case are also easy to handle. QED
Now let $A$ and $B$ be two nonempty disjoint subsets of $X$. Let $C$ be the disjoint union, so that we have a natural mapping
$\iota: C \to \{0,1\}$
sending elements of $A$ to $0$ and elements of $B$ to $1$.
We define an equivalence relation on $A$ with $a_0 \equiv a_1$ if the closed interval $[a_0,a_1] \cup [a_1,a_0]$ in $X$ has an empty intersection with $B$. The same logic would give us an equivalence relation on $B$, and this gives us a partition of the set $C$ itself, with the surjective mapping
$\pi: C \to \hat C$
$\rho \mapsto \bar \rho$
The function $\iota: \hat C \to \Bbb Z_2$ is also defined here.
Proposition 2: The ordering on $X$ defines an ordering on $\hat C$.
In general, $\hat C$ can have a maximum or a minimum point, but in the remainder of this section we impose the following constraint:
$\tag 1 \hat C \text { has no maximum point or minimum point}$
Proposition 3: The successor function $\sigma$ is an order preserving automorphism of $\hat C$ satisfying
$\quad \iota(\sigma(\bar \rho)) = \iota(\bar \rho) + 1$.
Proof Sketch: First show that $\sigma$ is well defined, etc..
For each $\bar \rho$, define
$H_{\bar \rho} = \{c \in C \; | \, \iota(c) = \iota(\bar \rho) \text{ and }c \le \rho \text { for all } \rho \in \bar \rho \}$
and
$K_{\bar \rho} = \{c \in C \; | \, \iota(c) = \iota(\bar \rho) + 1 \text{ and }c \ge \rho \text { for all } \rho \in \sigma (\bar \rho) \}$
When $H_{\bar \rho} \lt K_{\bar \rho}$ in $X$ are separated as described by Theorem 1, we can assign the disjoint open supersets
$R_{H_{\bar \rho}}$ and $R_{K_{\bar \rho}}$
Proposition 4: The intersection
$R_{K_{\bar \rho}} \bigcap R_{H_{\sigma(\bar \rho)}}$
meets exactly one of the sets $A$ or $B$.
So we define the 'even(=$A$)/odd(=$B$)' open sets
$U^0_{\bar \rho} = R_{K_{\bar \rho}} \bigcap R_{H_{\sigma(\bar \rho)}}$ superscript $0$ means it covers part of $A$
or
$U^1_{\bar \rho} = R_{K_{\bar \rho}} \bigcap R_{H_{\sigma(\bar \rho)}}$ superscript $1$ means it covers part of $B$
Theorem 5: If $A$ and $B$ are separated sets in $X$ there exist disjoint open supersets
$\quad A \subset U^0$
$\quad B \subset U^1$
Proof: Exercise.
The reader can prove Theorem 5 with the constraint (1) removed. A simple argument can handle the case when one of the sets $A$ or $B$ is all that remains as we move along $X$ to the $\pm \infty$ tails.
The proof by Cater in the following paper is particularly simple. But he shows (with little additional effort) that LOTS are collectionwise normal.
Cater, Frank S. (2006). "A Simple Proof that a Linearly Ordered Space is Hereditarily and Completely Collectionwise Normal". Rocky Mountain Journal of Mathematics. 36 (4): 1149–1151. doi:10.1216/rmjm/1181069408. ISSN 0035-7596. Zbl 1134.54317.
