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 $a \prec p$.

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 $Y \cap {\bar\downarrow}a \in \mathcal F$.

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 a filter on a set is proper.

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

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 $p \prec b$.

Either $Y \cap {\downarrow} b \in \mathcal F$ or $Y \cap {\bar \uparrow} b \in \mathcal F$.

Suppose, for the sake of contradiction, that $Y \cap {\bar \uparrow} b \in \mathcal F$.

Let $b' = \inf \left({ Y \cap {\bar \uparrow} b}\right)$.

Since $b$ is a lower bound of $Y \cap {\bar \uparrow} b$, $b \preceq b'$ by the definition of infimum.

Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity.

By the definition of $b'$ and the definition of $f$:
 * $b' \in f \left({\mathcal F}\right)$

But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $f \left({\mathcal F}\right)$