Compact Subspace of Linearly Ordered Space

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

Let $Y \subseteq X$ be a non-empty 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
Let $S$ be a nonempty subset of $Y$.

By Compact Subspace of Linearly Ordered Space/Lemma, $S$ has a supremum $k$ in $Y$.

Aiming for a contradiction, suppose that $S$ has an upper bound $b$ in $X$ such that $b \prec k$.

Let $\mathcal A = \{{ {\downarrow}s:s \in S }\} \cup \{{ {\uparrow}b }\}$

Then $\mathcal A$ is an open cover of $Y$.

But $\mathcal A$ has no finite subcover, contradicting the fact that $Y$ is compact.

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

Also see

 * Heine–Borel Theorem: Dedekind-Complete Space
 * Connected Subspace of Linearly Ordered Space