Is the support of a compactly supported function on $\Omega$ a proper subset of $\Omega$? Let $d\in\mathbb N$ and $\Omega\subseteq\mathbb R^d$ be open. Is there some continuous $\phi:\Omega\to\mathbb R$ with compact support $\operatorname{supp}$ and $\operatorname{supp}\phi=\overline{\Omega}$?
 A: Let $\Omega$ be the open unit ball and let $\phi(x) = 1$.  Then by your definition of support, $\mathrm{supp\, }\phi = \overline \Omega$.
A: Given a topological space $\Omega$, e.g., a subset of ${\mathbb R}^d$ given the relative topology, and a  function $\phi:\>\Omega\to{\mathbb R}$ the support of $\phi$ is by definition the closure (in $\Omega$) of the set of $x\in\Omega$ where $\phi(x)\ne 0$:
$${\rm supp}\>\phi=\overline{\{x\in\Omega\>|\>\phi(x)\ne0\}}\ \subset\Omega\ ,$$
and is a subset of the given space $\Omega$. Note that $\Omega=\bar\Omega$ in this context.
Example: If $\Omega\subset{\mathbb R}^2$ is the open unit disc, and $\phi(x)=1-|x|$, then ${\rm supp}\>\phi=\Omega$, since $\phi$ is nonvanishing on $\Omega$. If $\Omega\subset{\mathbb R}^2$ is the closed unit disc, and $\phi(x)=1-|x|$, then ${\rm supp}\>\phi=\Omega$ as well, since no point on the unit circle has a neighborhood on which $\phi$ vanishes identically.
All of this has nothing to do with the fact that $\Omega$ might be a subspace of some larger space $X$. If the latter is the case then $\Omega$ has a closed hull in $X$. But even if this hull is a compact set larger than $\Omega$ this would not make the support $S$ of $\phi$, as defined in $(1)$, compact if $S$ didn't have this property right from the outset.
A: 
For any open $\Omega$, does there exist a continuous $\phi:\Omega \to \mathbb{R}$ with compact support equal to $\overline{\Omega}$?

No. Consider $\Omega = \mathbb{R}^d$. Then the support of any compactly supported function will never equal $\overline{\mathbb{R}^d} = \mathbb{R}^d$, which is not compact.

More generally, we have the inclusions $\operatorname{supp}\phi \subset \Omega \subset \overline{\Omega}$. The latter inclusion is strict if and only if $\Omega$ isn't closed. Therefore the only way to possibly have $\operatorname{supp} \phi = \overline{\Omega}$ is for $\Omega$ to be both open and closed, i.e., $\Omega = \emptyset$ or $\Omega = \mathbb{R}^d$.
If $\Omega = \emptyset$, the empty function trivially satisfies your request.
If $\Omega = \mathbb{R}^d$, there is no function that satisfies your request, as I've shown above.
