If $X$ is an order topology and $Y \subset X$ is closed, do the subspace topology and order topology on $Y$ coincide?
No. For example, let $X$ be $\mathbb{Q}$ with its standard ordering and let $Y$ be $([0,\sqrt{2}] \cap \mathbb{Q})\cup \{2\}$. Then $Y$ is closed in $X$ but the two topologies do not coincide: $2$ is isolated in the subspace topology but not in the order topology.