Order Topology on Convex Subset is Subspace Topology

Theorem
Let $(S,\le,\tau)$ be a linearly ordered space.

Let $A \subseteq S$ be a convex set in $S$.

Let $\upsilon$ be the order topology on $A$.

Let $\tau'$ be the $\tau$-relative subspace topology on $A$.

Then $\upsilon = \tau'$.

Proof
Each open ray in $A$ is trivially the intersection of an open ray in $S$ with $A$. Since the open rays in $A$ form a subbase for $\upsilon$, $\upsilon \subseteq \tau'$.

Suppose now that $p \in S$.

If $p \in A$, then $A \cap {\uparrow_S}p = {\uparrow_A}p$, so $A \cap {\uparrow_S}p \in \upsilon$.

Suppose instead that $p \notin A$.

Let $x \in A \cap {\uparrow_S}p$ (if there is no such $x$, then the set is empty, and hence in $\upsilon$).

Since $A$ is convex in $S$, no element of $A$ precedes $p$.

Thus $A \cap {\uparrow_S}p = A$, which is in $\upsilon$ by the definition of a topology.

We have shown that the intersection of each upward ray in $S$ with $A$ is in $\upsilon$. The same holds for downward rays by the same argument.

Since each element of a subbase of $\tau'$ is in $\upsilon$, $\tau' \subseteq \upsilon$.