Heine-Borel Theorem/Dedekind Complete Space
Let $Y$ be a compact subspace of $T$.
it follows that $Y$ is closed in $T$.
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 Closed Set in Linearly Ordered Space, $b$ is not a supremum of $S$, a contradiction.
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.