For a space $X$ and a linear order $<$ on $X$, we can define a topology $\tau$ by the basic open sets $U_{a,b}=\{x\in X: a<x<b\}$. Now, let $A\subset X$. We can endow it with the subspace topology $\tau'$ in the usual way, by calling all $U\cap A$, where $U\in \tau$, open. Moreover, we can consider the restriction of the order $<$ to $A$, and endow $A$ with the topology of open intervals by $U_{c,b}=\{x\in A: c<x<b\}$ (call this topology $\tau''$).
My question is: do we have that $\tau'=\tau''$?
Let $X=\mathbb{R}$ and $A=[0,1]\cup (2,3]$, then $(\frac{1}{2},1] = (\frac{1}{2},2)\cap A$ and so it is open in the subspace topology. But it's not open in the order topology.