Compact Subspace of Linearly Ordered Space

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

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

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$ iff both of the following hold:
 * $(1):\quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
 * $(2):\quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Forward Implication
$(1)$ holds by Compact Subspace of Linearly Ordered Space/Lemma.

A linearly ordered space is Hausdorff by Order Topology is Hausdorff.

A compact subset of a Hausdorff space is closed by Compact Subspace of Hausdorff Space is Closed.

Thus $Y$ is closed in $X$.

Suppose for the sake of contradiction that for some nonempty $S \subseteq Y$, $\sup S \notin Y$.

Then for every $x < \sup S$, $\left({{x}\,.\,.\,{\sup S}}\right] \cap Y \ne \varnothing$.

Thus $\sup S$ is a limit point of $Y$ not contained in it, contradicting the fact that $Y$ is closed.

Thus $\sup S \in Y$ for each nonempty $S \subseteq Y$.

A similar argument proves the corresponding statement for infima, so $(2)$ holds.