Heine-Borel Theorem/Dedekind Complete Space

Theorem
Let $(X, \preceq, \tau)$ be a Dedekind-complete linearly ordered space.

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

Then $Y$ is compact iff $Y$ is closed and bounded in $X$.

Forward implication
Follows directly from Compact Subspace of Linearly Ordered Space/Lemma, Order Topology is Hausdorff, and Compact Subspace of Hausdorff Space is Closed.

Reverse Implication
Let $Y$ be a closed, bounded subset of $X$.

Let $S$ be a non-empty subset of $Y$.

Since $Y$ is bounded and $S \subseteq Y$, $S$ is bounded.

Since $X$ is Dedekind complete, $S$ has a supremum and infimum in $X$.

We will show that $\sup S, \inf S \in Y$.

Suppose for the sake of contradiction that $\sup S \notin Y$.

By Heine-Borel Theorem/Dedekind-Complete Space/Lemma, there is an $a \prec \sup S$ such that
 * ${\uparrow}a \cap S = \varnothing$.

But then $a$ is an upper bound of $S$ and $a \prec \sup S$, contradicting the definition of supremum.

Thus $\sup S \in Y$.

A similar argument shows that $\inf S \in Y$.

Thus by Compact Subspace of Linearly Ordered Space, $Y$ is compact.