Compact Subspace of Linearly Ordered Space/Reverse Implication

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

Let $Y \subseteq X$ be a nonempty subset of $X$.

Suppose that for every non-empty $S \subset Y$, $S$ has a supremum and an infimum in $X$, and $\sup S, \inf S \in Y$.

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$.

Proof
Let $\mathcal F$ be an ultrafilter on $Y$.

For $S \in \mathcal F$, let $f(S) = \inf S$.

Let $p = \sup f(\mathcal F)$.

Then $\mathcal F$ converges to $p$:

Upward rays
Let $a \in X$ with $a < p$.

Since $\mathcal F$ is an ultrafilter, either $Y \cap {\uparrow}a \in \mathcal F$ or $Y \cap {\bar\downarrow}a \in \mathcal F$.

Suppose for the sake of contradiction that ${\bar\downarrow}a \in \mathcal F$.

For each $S \in \mathcal F$:
 * $S \cap {\bar\downarrow}a \in \mathcal F$ because an ultrafilter is a filter.
 * $S \cap {\bar\downarrow}a \ne \varnothing$ because an ultrafilter is a non-trivial filter.

By the definition of supremum, there exists an $S \in \mathcal F$ such that $a < \inf S$.

By the definition of infimum, $S \cap {\bar\downarrow}a = \varnothing$, a contradiction.

Thus $Y \cap {\uparrow}a \in \mathcal F$.

Downward rays
Let $b \in X$ with $p < b$.