Separated Subsets of Linearly Ordered Space under Order Topology/Lemma

Lemma
Let $T = \left({S, \preceq, \tau}\right)$ be a linearly ordered space.

Let $A$ and $B$ be separated sets of $T$.

Let $A^*$ and $B^*$ be defined as:
 * $A^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in A, \left[{a \,.\,.\, b}\right] \cap B^- = \varnothing}\right\}$
 * $B^* := \displaystyle \bigcup \left\{ {\left[{a \,.\,.\, b}\right]: a, b \in B, \left[{a \,.\,.\, b}\right] \cap A^- = \varnothing}\right\}$

where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.

Then:
 * $(1): \quad A \subseteq A^*$


 * $(2): \quad B \subseteq B^*$


 * $(3): \quad A^* \cap B^* = \varnothing$

Proof
Let $a \in A$.

Then:

Similarly, $B \subseteq B^-$.

$A^* \cap B^* \ne \varnothing$.

Then:
 * $\exists p: p \in A^* \cap B^*$

Hence:
 * $\exists a, b \in A, c, d \in B: p \in \left[{a \,.\,.\, b}\right] \cap \left[{c \,.\,.\, d}\right]$

But because $A$ and $B$ are separated sets:
 * $c, d \notin A$

and:
 * $a, b \notin B$

and so:
 * $\left[{a \,.\,.\, b}\right] \cap \left[{c \,.\,.\, d}\right] = \varnothing$

Thus $p \notin \left[{a \,.\,.\, b}\right] \cap \left[{c \,.\,.\, d}\right]$

It follows by Proof by Contradiction that $A^* \cap B^* = \varnothing$.