Holes of a compact set in $\mathbb{R}^n$ that do not contain holes of a larger open set Let $K$ be a compact subset of an arbitrary open set $\Omega\subset \mathbb{R}^n.$ It is said that a connected component $W$ of $\Omega\setminus K$ is $\Omega$-bounded if $\overline{W}$ is a compact subset of $\Omega$.
Heuristically speaking, the $\Omega$-bounded components of $\Omega\setminus K$ are  the "holes" of $K$ that do not contain "holes" of $\Omega$. Now, I want to prove this, but I am lost. Topology is not my usual work field. 
Specifically, I want to prove that if $W$ is an $\Omega$-bounded component of $\Omega\setminus K$ and $H$ is the component of $\mathbb{R}^n\setminus K$ that contains $W,$ then $H\subset \Omega.$ Additionally, is it possible to prove that $H=W$?
 A: Yes, $W=H$, so in particular $H \subset \Omega$.
(All uses of closure, boundary, open, closed, complement, etc, are relative to  $\mathbb{R}^n$ unless otherwise stated.)
First, note that for any open subset $U \subset \mathbb{R}^n$, a connected subset of $U$ is a component of $U$ iff it is open and relatively closed in $U$.  This follows from the local connectedness of $\mathbb{R}^n$.
Claim. $\partial W \subset K$.
Proof.  Let $x \in \partial W$.  Clearly $x \notin W$ since $W$ is open.  Since $W$ is $\Omega$-bounded, we have $x \in \Omega$.  If $x \notin K$, then $x$ is in some other component $W'$ of $\Omega \setminus K$.  But $W'$ is open and hence $W'$ is not disjoint from $W$, which is absurd. $\Box$
Now since $W$ is open, the claim implies that $W = \overline{W} \setminus \partial W = \overline{W} \cap K^c$.  So $W$ is relatively closed in $K^c$.  Being connected and relatively clopen means that $W$ is a connected component of $K^c$, hence it equals the component $H$ which contains it.
Note we didn't use the compactness of $K$ (closed would suffice) nor of $\overline{W}$.
