$\Omega$ open and $K$ compact $\Rightarrow B_\varepsilon(0)+K\subset\Omega$ for $\varepsilon$ small enough. Let $\Omega\subset\mathbb{R}^N$ be an open bounded set and $K\subset\Omega$ compact.
I'm trying to prove that for $\varepsilon$ small enough we have $(B_\varepsilon(0)+K)\subset\Omega$.
Consider the case $N=1$, $\Omega=(a,b)$ and $K=[c,d]$. If we take $\varepsilon<\min\{b-d,c-a\}$ then
$$\begin{align*}x\in(B_\varepsilon(0)+K)&\Rightarrow x=y+z\text{ with }y\in (-\varepsilon,\varepsilon)\text{ and }z\in[c,d]\\
&\Rightarrow-\varepsilon+c < x<\varepsilon+d\\
&\Rightarrow a < x<b\\
&\Rightarrow x\in \Omega
\end{align*}$$
so that the inclusion holds.
How to deal with the general case?

For each $x \in K \subseteq \Omega$, there is an open ball $B(x, r_x)$ around $x$, and contained in $\Omega$. But then $$\{B(x, r_x) : x \in X\}$$
forms an open cover of $K$, so....
