Proof of Theorem 9.2 in Brezis' Functional Analysis, Sobolev spaces and Partial Differential Equations I am currently reading Brezi's Functional Analysis, Sobolev spaces and Partial Differential Equations. I am somehow stuck in the proof of Theorem 9.2 where the setup is the following:

$\Omega \subset \mathbb{R}^N$ is open and $\omega \subset \subset \Omega$, i.e. $\overline{\omega} \subset \Omega$. Let $\alpha \in C_c^1(\Omega)$ (i.e. continuous and compactly supported in $\Omega$) such that $\alpha = 1$ on a neighbourhood of $\omega$ Let $\overline{\alpha}$ be the function $\alpha$ extended to be zero on $\Omega^c$. Let $B(0,1/n)$ denote the open ball of radius $1/n$ centered at $0$.

Now the question: How do I see that I have
$$\overline{B(0,1/n) + \text{supp}(1-\overline{\alpha})} \subset (\omega)^c$$
for sufficiently large $n$? 
 A: The claim is equivalent to saying that 
$$\operatorname{dist}(\operatorname{supp}(1-\overline{\alpha}), \overline{\omega})>0 \tag1$$
Indeed, if (1) holds, then taking $n$ such that $1/n$ is less than the distance in (1), we get the conclusion. 
The set $\operatorname{supp}(1-\overline{\alpha})$ is the closure of the open set $\{x:\overline{\alpha}(x)\ne 1\}$. The latter set is disjoint from a neighborhood of $\omega$ by the assumption "$α=1$ on a neighbourhood of $ω$". Therefore, $\operatorname{supp}(1-\overline{\alpha})$ is disjoint from $\overline{\omega}$. 
It remains to recall:  A and B disjoint, A compact, and B closed implies there is positive distance between both sets. Specifically, $\overline{\omega}$ is compact* and $\operatorname{supp}(1-\overline{\alpha})$ is closed.
(*) Why? Because the notation $\omega \subset\subset \Omega$ (often written $\omega\Subset \Omega$) means that $\overline{\omega}$ is a compact subset of $\Omega$. For example, it does not apply to the sets 
$$
\Omega=\{(x, y)\in\mathbb{R}^2 : y > 0\},\quad 
\omega=\{(x, y)\in\mathbb{R}^2 : y > e^x\}, 
$$
even though $\overline{\omega}\subset\Omega$ here. 
Alternatively (if you don't buy my interpretation of notation), note that $\overline{\omega}\subset\operatorname{supp}\alpha$, and the latter is compact. 
