I read on another thread here that the set $\{0\}$ is open in $\{0,1\}$, with $\{0, 1\}$ a subset of $\textbf{R}$. This makes sense to me b/c $\exists$ an open set in $\textbf{R}$, say, $(-1,1)$ s.t. $\{0\}$ = $(-1,1) \cap \{0, 1\}$. But then since $\{0\}$ = $[-1,.5] \cap \{0, 1\}$, does that also imply that $\{0\}$ is closed in $\{0, 1\}$ as well?
Thank you for any clarification.
Yes, that is correct.
Alternatively, note that just as $\{0\}$ is open in that set, so too is $\{1\}$. Therefore the complement of $\{1\}$ is closed, but its complement is just $\{0\}$.
This suggests a more general proof.
If it is closed, its complement in the subspace is open, so its complement can be written $$U^c=O\cap S$$ where $O$ is an open set and $S$ is the subspace. Then $$O^c\cap S=(O\cap S)^c=U$$ so $U$ is the intersection of a closed set with $S$ (this uses the properties of how a subspace intersects with other sets).
Conversely, suppose that we have a set $U$ which is an intersection $$U=C\cap S$$ where $C$ is closed. Then its complement (in the subspace, note) is $$U^c=(C\cap S)^C=S\cap (C^c)$$ is, by definition, an open set in the subspace topology. So $U$ has open complement, and is closed itself. Hence we have proved the claim.