Complete Linearly Ordered Space is Compact

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

Let $(X, \preceq)$ be a complete lattice.

Then $(X, \tau)$ 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 $(X, \preceq)$ is complete.

Let $C$ be the set of all $x \in X$ such that a finite subset of $\mathcal A$ covers $[m \,.\,.\, x]$.

$C$ 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 $U = (a \,.\,.\, b)$, $U = {\dot\uparrow} a$, or $U = {\dot\downarrow} b$.

Suppose that $U = (a \,.\,.\, b)$.

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 $[m \,.\,.\, x]$.

Then $\mathcal F \cup \{U,V\}$ covers $[m \,.\,.\, b]$, 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 $[m \,.\,.\, b]$ is covered by $\{ U, V \}$, 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 $[m \,.\,.\, x]$.

Thus $\mathcal F \cup \{ U \}$ is a finite subcover of $A$.