$X$: ordered space
$Y$: subset of $X$
If $Y$ is not a convex subset of $X$, the order topology of $Y$ and the subspace topology of $Y$ need not be the same.
Example: If $X=\mathbb{R}, Y=[0,1) \cup \{2\}$, then $\{2\}$ is open in $Y$ in the subspace topology but not open in the order topology.
But is it true that the subspace topology is always finer than the order topology? This is my question.
Yes. If $\tau_Y$ is the subspace topology on $Y$, and $\tau_{\le}$ is the order topology on $Y$, then $\tau_{\le}\subseteq\tau_Y$. Suppose that $u,v\in Y$ with $u<v$: then
$$\{y\in Y:u<y<v\}=(u,v)\cap Y\in\tau_Y\;,$$
so the natural base for $\tau_{\le}$ is a subset of $\tau_Y$, and therefore $\tau_{\le}$ itself is a subset of $\tau_Y$.
Added: If $Y$ has a maximum element $y^+$, we must also consider the sets of the form
$$\{y\in Y:u<y\le y^+\}$$
for arbitrary $u\in Y\setminus\{y^+\}$, but these occasion no real difficulty. If $y^+=\max X$, then
$$\{y\in Y:u<y\le y^+\}=(u,y^+]\cap Y\in\tau_Y\;,$$
and if not, we simply pick some $x\in X$ with $y^+<x$ and note that
$$\{y\in Y:u<y\le y^+\}=(u,x)\cap Y\in\tau_Y\;.$$
The case in which $Y$ has a least element is similar.
