Bounded Closed Intervals are Compact in the Order Topology iff the Order is Complete
# Statement
Let $(X, \leq)$ be a Total Ordering endowed with the Order Topology. Then for $a,b \in X$, $[a,b]$ is Compact If and Only If $(X, \leq)$ is a Complete Ordering.
# Proof
$(\Leftarrow)$: Let $a,b \in X$. $[a,b]$ is a Tightly Bounded Set. $[a,b]$ is also Closed since Closed Intervals are Closed. Recall Closed Subset of a Complete Space is Complete and A Nonempty Set is Compact in the Order Topology iff it is Tightly Bounded and Complete, so $[a,b]$ is Compact $\checkmark$.
$(\Rightarrow)$: Let $A \subset X$ be a Nonempty Bounded Set. Suppose $a \in A$ and $b \geq A$, and let $S := A \cap [a,b]$. We see $\sup\limits S \in S$ because $[a,b]$ is Compact and Compact Spaces are Complete. $S$ is a Cofinal Subset of $A$ (either $x < a$ or $x \in [a,b]$ for $x \in A$). Because Supremum of a Cofinal Subset is the Supremum of the Set, $\sup\limits A = \sup\limits S$. Therefore $\sup\limits A$ exists. Since $A$ was arbitrary, $(X, \leq)$ is a Complete Ordering $\checkmark$. $\blacksquare$