Closed Set in Linearly Ordered Space

Theorem
Let $\left({X,\preceq,\tau}\right)$ be a linearly ordered space.

Let $Y$ be a closed subset of $X$.

Let $S \subseteq Y$.

Let $b \in X \setminus Y$.

Let $b \in X$ be an upper bound of $S$.

Let $b \notin Y$.

Then there is an $a \in X$ such that $b \in {\uparrow}a$ and ${\uparrow}a\cap S = \varnothing$.

Proof
Since $Y$ is closed and $b \notin Y$, there must be an open interval or open ray $U$ containing $b$ that is disjoint from $Y$.

Since $b$ is an upper bound for $S$ and $S$ is not empty, $U$ cannot be a downward-pointing ray.

Thus it is either an open interval or an open ray.

Let $a$ be the left endpoint of $U$.

Then $a \prec b$.

Since $b \in U$, and $b$ is an upper bound of $S$, no element strictly succeeding all elements of $U$ can be in $S$.

By the above and the fact that $S \subseteq Y$, ${\uparrow}a \cap S = \varnothing$.