Let $X \subset \mathbb{R}$ and give $X$ its order topology. (When) is it true that $X$ is homeomorphic to some subspace of $Y \subset \mathbb{R}$?
Example: Let $X = [0,1) \cup \{2\} \cup (3,4]$, then $X$ is order isomorphic to $Y=[0,2]$ in a fairly obvious way, hence the order topologies on $X$ and $Y$ are homeomorphic. But, the order topology on $Y$ agrees with the subspace topology on $Y$, so this particular $X$ is indeed homeomorphic to a subspace of $\mathbb{R}$.
Example: Let $X$ equal the set of endpoints of the Cantor set. The order on $X$ is a bit strange (it's order isomorphic to $\mathbb{Q} \times \{0,1\}$ in the dictionary order with a largest and smallest element adjoined), so one might expect the order topology on $X$ to be strange, but... $X$ is clearly 2nd countable under the order topology (there are countably many points, hence countably many intervals) and any order topology is regular, so $X$ is metrizeable by Urysohn. It is also easy to see the order topology on $X$ has no isolated points so, by a theorem of Sierpinski (any countable metric space without isolated points is homeomorphic to $\mathbb{Q}$), the order topology on $X$ is homeomorphic to the subspace $Y = \mathbb{Q}$ of $\mathbb{R}$.
For any countable $X$, the argument above shows the order topology on $X$ is metrizeable. In fact, I think the order topology on any $X \subset \mathbb{R}$ is metrizeable by this theorem of Lusin (an order topology $X$ is metrizeable if and only if the diagonal in $X \times X$ is a $G_\delta$ set). But this doesn't preclude the possiblity of $X$ weird enough to not be homeomorphic to any subspace of $\mathbb{R}$. I'm not sure how one would arrange this though. If it can be arranged with $X$ countable there would need to be many isolated points to avoid Sierpinski's result for instance...
Advertisement: Brian's positive answer below actually proves something even better. Namely:
Theorem: Let $X \subset \mathbb{R}$. Then there is an order isomorphism (hence homeomorphism of order topologies) $\varphi$ from $X$ to a new set $Y \subset \mathbb{R}$ whose order topology and subspace topology coincide.
Roughly, approach is to take countable set $D \subset X$ such that each point in $X \setminus D$ is a 2-sided limit (topologically) of points in $D$. Using countability of $D$, one can find an order embedding $\varphi$ of $D$ into $\mathbb{Q}$ such that $\varphi(D)$ has no "topological gaps" that are not also "order gaps" (this is the property which makes the subspace and order topologies coincide). The final step is to extend $\varphi$ to the rest of $X$ and check nothing goes awry in doing so.