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Here is Example 2, Sec. 16 in the book Topology by James R. Munkres, 2nd edition:

Let $Y$ be the subset $[0, 1) \cup \{ 2 \}$ of $\mathbb{R}$. In the subspace topology on $Y$ the one-point set $\{ 2 \}$ is open, because it is the intersection of the open set $\left( \frac{3}{2}, \frac{5}{2} \right)$ with $Y$. But in the order topology on $Y$, the set $\{ 2 \}$ is not open. Any basis element for the order topology on $Y$ that contains $2$ is of the form $$ \left\{ \ x \ | \ x \in Y \ \mbox{ and } \ a < x \leq 2 \ \right\}$$ for some $a \in Y$; such a set necessarily contains points of $Y$ less than $2$.

I think I've properly understood this example.

Now my question is, can we find a set in the order topology on $Y$ that does not belong to the subspace topology that $Y$ inherits from the standard topology on $\mathbb{R}$?

My feeling is that this is not possible.

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  • $\begingroup$ Digression: Although the subspace topology on $Y$ is not its order topology, the subspace $Y$ is homeomorphic to an ordered space :Let $f(x)=x$ for $x\in [0,1)$ and $f(2)=-2.$... On the other hand the subspace topology on $(0,1)\cup \{2\}$ cannot be generated by any linear order. $\endgroup$ – DanielWainfleet May 9 '17 at 18:46
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Hint: you can actually write the possible forms of the open sets of $Y$ under the order topology, now try to find an open set in R that will induce the same types of open set under the subspace topology(easy to find), so ans is no!

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On a subset $S$ of an ordered space $(T,<),$ the induced order topology is always a subset of the subspace topology.

A sub-base $B$ for the induced order topology on $S$ is all sets of the form $(\leftarrow,s)_S=\{u\in S:u<s\},$ or $(s,\to)_S=\{u\in S:u>S\},$ (over all $s\in S$), or $S$ itself. Each member of $B$ is open in the subspace topology as it it the intersection of $S$ with an open set of $T$. E.g. $(s,\to)_S=S\cap \{t\in T:t>s\}$ when $s\in S.$

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