can someone prove the following statement: Let X is an ordered set with the order topology. Let Y be a convex subset of X. Then the order topology on Y is the same as the subspace topology on Y.
intuitively it's kind of obvious but i can't prove it...
One inclusion always holds: suppose $a \in Y$ and consider a set of the form $(\leftarrow, a)_Y$ or $(a, \rightarrow)_Y$ (the subscript denotes that we take the order restricted to $Y$). These sets form a subbase for the order topology on $Y$, by definition.
Then $$\left(\leftarrow, a\right)_Y = \left\{y \in Y: y <_Y a \right\} = \left\{y \in X: y <_X a \right\} \cap Y$$
so this set is open in the subspace topology on $Y$ w.r.t. the order topology $\mathscr{T}_X$ on $X$. The same holds for the other type of subbasic open set. So $\mathscr{T}_{\!<_{Y}} \subseteq (\mathscr{T}_X) |_Y$.
So suppose we have $O \cap Y$ which is open in $(\mathscr{T}_X) |_Y$, where $O \in \mathscr{T}_X$ is basic open (it's enough to consider base sets).
So one case (the most common one) is that $O = (a,b)$, where $a < b$ are both in $X$. Suppose $y \in Y \cap (a,b)$. we want to show it is an interior point in $\mathscr{T}_{\!<_{Y}}$ of $Y \cap (a,b)$.
Some subcases: suppose there is some $z \in Y$ such that $ z \in (a,y)$. Then define $y_L = z$. If this is not the case, there can be some $z \in Y$ with $z \le a$. In that case, as $Y$ is order convex (!) and $z \le a < y$, we define $y_L = a \in Y$. Note that in either case $y_L \in Y$ and $a \le y_L < y$. In the remaining case, $y = \min(Y)$.
For the right side we make similar distinctions and get either that $y = \max(Y)$ or that there is some $y_R \le b$ with $y < y_R \in Y$.
If we have $y_L, y_R$ both, then $y \in (y_L, y_R)_Y$ and the latter is open in the order topology on $Y$ (both endpoints are in $Y$) and this sits in $O \cap Y$.
If we have $y = \min(Y) = \max(Y)$, we're done anyway, as then $Y$ is a singleton and has only one topology. Of we have (say) $y = \min(Y)$ and $y_R$ then $[y, y_R)_Y$ is the required $Y$-order open set, or else $(y_L, y]_Y$ in the $\max(Y)$ case. So in all cases we are done.
The remaining cases (quite similar, bit easier) are for the cases $O = [\min(X),a)$ and $O = (a,\max(X)]$ (if those min or max exist, but we need to consider all cases). You can check those yourself.
A classic example of why we need something extra on $Y$: $Y = [0,1) \cup \{2\}$, which is not order convex, in the usual real order, has $2$ as an isolated point in the subspace topology (witnessed by $(1,3) \cap Y = \{2\}$, but $2$ is the maximum in the restricted order topology $2$ so only has neighbourhoods of the form $(a,2]_Y$ where $a \in Y, a < 2$, so in particular $a < 1$ and so there are always infinitely many points in any neighbourhood of $2$ in that topology. So there the subspace topology is strictly bigger.
