Complete Linearly Ordered Space is Compact
Theorem
Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.
Let $\left({X, \preceq}\right)$ be a complete lattice.
Then $\left({X, \tau}\right)$ is compact.
Proof
By Compactness from Basis, it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover.
Let $\mathcal A$ be an open cover of $X$ consisting of open rays and open intervals.
Let $m = \inf X$. This infimum exists because $\left({X, \preceq}\right)$ is complete.
Let $C$ be the set of all $x \in X$ such that a finite subset of $\mathcal A$ covers $\left[{m \,.\,.\, x}\right]$.
$C$ is non-empty because $m \in C$.
Let $s = \sup C$.
Since $\mathcal A$ covers $X$, there is a $U \in \mathcal A$ such that $s \in U$.
Then we must have $U = \left({a \,.\,.\, b}\right)$, $U = {\dot\uparrow} a$, or $U = {\dot\downarrow} b$.
Suppose that $U = \left({a \,.\,.\, b}\right)$.
Let $V \in \mathcal U$ contain $b$.
Then by the definition of supremum, there is an $x \succ a$ such that there is a finite $\mathcal F \subseteq \mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$.
Then $\mathcal F \cup \left\{{U, V}\right\}$ covers $\left[{m \,.\,.\, b}\right]$, contradicting the fact that $s$ is an upper bound of $C$.
Suppose next that $U = \dot\downarrow b$.
Then for some $V \in \mathcal A$, $b \in V$.
Then $\left[{m \,.\,.\, b}\right]$ is covered by $\left\{{U, V}\right\}$, contradicting the fact that $s$ is the supremum of $C$.
Thus $U = \dot\uparrow a$.
By the definition of supremum, $a$ is not an upper bound of $C$.
So there is an $x \succ a$ such that there is a finite subset $\mathcal F$ of $\mathcal A$ that covers $\left[{m \,.\,.\, x}\right]$.
Thus $\mathcal F \cup \left\{{U}\right\}$ is a finite subcover of $A$.
$\blacksquare$
Sources
1955: John L. Kelley: General Topology: $\S 1$: Problem $\text I$, $\S 5$: Problem $\text C$