Do test functions have smooth extensions? Let $\Omega\subset\mathbb R^d$ be open. A smooth function $\phi\colon \Omega\to\mathbb R$ is called a test function on $\Omega$ if its support is compact and inside $\Omega$. I expect that the function
\begin{align}
\Phi\colon\mathbb R^d&\to\mathbb R\\
x&\mapsto\begin{cases}\phi(x)&\text{if }x\in\Omega\\
0&\text{else}\end{cases}
\end{align}
is a test function on $\mathbb R^d$, but there is one issue. We need to show that $$U:=\{x\in\mathbb R^d:\Phi\text{ is smooth in }x\}=\mathbb R^d$$ but it is not clear to me why $\partial\Omega\subset U$.
My attempt
My guess is that each $x\in\partial\Omega$ has a neighbourhood where $\Phi$ is zero, because if this is not the case for some $x\in\partial\Omega$, then this $x$ is inside the support of $\Phi$, but by the above definition the support of $\phi$ - which equals the support of $\Phi$ - is inside $\Omega$. However, I don't know how to make this rigorous, since I have not attended a topology lecture. Also, my guess might be wrong. So your help would be appreciated.

Thanks to the hint given in the comments, I was able to complete the proof. We can show that for each $x\in\partial\Omega$ there is an $r>0$ s.t. $\Phi= 0$ on the open ball $B(x,r)=\{y\in\mathbb R^d:|x-y|<r\}$. In my question I suggested to prove the weak existence of such an $r$, i.e. $\lnot\forall_{r>0}\lnot\forall_ {y\in B(x,r)}\Phi(y)=0$, but we can actually give a constructive proof.
Proof:
Let $K$ be the support of $\phi$ and consider some $x\in\partial\Omega$. Since the function $K\ni y\mapsto|x−y|$ is continuous and defined on a compact set, there is some $a\in K$ s.t. $|x−a|≤|x−y|$ for all $y\in K$. Of course, $0<|x−a|=:r$ since $\partial\Omega$ and $\Omega$ are disjoint. Thus, $K\cap B(x,r)=\emptyset$ and $\Phi=0$ on $B(x,r)$.
