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

Then $b$ is not a supremum of $S$.

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 $U$ is either an open interval $\left({{a}\,.\,.\,{q}}\right)$ or an upward-pointing open ray ${\uparrow}a$.

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 $S \subseteq Y$, ${\uparrow}a \cap S = \varnothing$, so $a$ is an upper bound of $S$.

Since $a \prec b$, $b$ is not a supremum of $S$.