Suppose we have a bounded open set $O \subset R^n$. Let $F$ be a closed subset in the subspace $O$ which is bounded away from the boundary of $O$. That is, there exists $\delta > 0$, for every $y \in F$, we have \begin{align*} \|x - y\| > \delta, \end{align*} for all $x \in \partial O$ (boundary of $O$).
I am thinking whether this will imply $F$ is compact. Boundedness is not an issue and we only need to make sure $F$ is closed in $\mathbb R^n$. Intuitively, I think the set that is closed in subspace $O$ but not in $\mathbb R^n$ must have the property: $\partial F \cap \partial O \neq \emptyset$, where we consider $F$ and $O$ as both subsets of $\mathbb R^n$. I made up this condition and wonder whether we can prove this? If this condition is not correct, what kind of reasonable condition would guarantee $F$ is compact?
Let $(x_n)_{n\in\mathbb N}$ be a convergent sequence of elements of $F$ and let $x$ be its limit. I will prove that $x\in F$. If $x\in O$, then $x$ in $F$, since $F$ is a closed subset of $O$. And if $x\not\in O$, then $x\in\partial O$, since $(x_n)_{n\in\mathbb N}$ is a sequence of elements of $O$. But this is impossible, since $F$ is bounded away from the boundary of $O$.
Since each convergent sequence of elements of $F$ converges to an element of $F$, $F$ is closed (in $\mathbb{R}^n$). And obviously (as you wrote) $F$ is bounded. Therefore, $F$ is compact.