Heine-Borel iff Dedekind Complete

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

Then $X$ is Dedekind complete every closed, bounded subset of $X$ is compact.

Proof
The forward implication follows from Heine-Borel Theorem: Dedekind-Complete Space.

Suppose that $X$ is not Dedekind complete.

Then $X$ has a non-empty subset $S$ with an upper bound $b$ in $X$ but no supremum in $X$.

Let $a \in S$ and let $Y = {\bar\downarrow}S \cap {\bar\uparrow} a$.

$Y$ is nonempty and bounded below by $a$ and above by $b$.

$Y$ is closed in $X$
Let $x \in X \setminus Y$.

Then $x \prec a$ or $x$ strictly succeeds every element of $S$.

If $x \prec a$, then $x \in {\downarrow} a \subseteq X \setminus Y$.

If $x$ strictly succeeds each element of $S$, then it is an upper bound of $S$.

Since $S$ has no supremum in $X$, it has an upper bound $p \prec x$.

Then $x \in {\uparrow p} \subseteq X \setminus Y$.

$Y$ is not compact
Let $\mathcal A = \left\{ { {\downarrow} s: s \in S }\right\}$.

Then $\mathcal A$ is an open cover of $Y$ with no finite subcover.