I'm trying to show that if $G$ is an open set and $K$ is a compact set with $K \subset G$, show that there is a $\delta > 0$ such that $\{x: \textrm{dist}(x, K) < \delta \} \subset G$.
My current attempt involves using a proof by contradiction. Letting $\delta = 1/n$ we can generate a sequence $\{x_n\}$ of points that are not in $G$. I think there should be way to show that this sequence converges to a point in $K$, leading to a contradiction, but I'm not entirely sure how to proceed.
Let $d$ be the metric. Assume $K\ne \emptyset.$
For each $y\in K$ let $r_y>0$ such that the open ball $B_d(y,r_y)$ is a subset of $G.$
Now $\{B_d(y,r_y/3):y\in K\}$ is an open cover of $K,$ and $K$ is compact, so there exists a non-empty finite $L\subset K$ such that $\{B_d(y,r_y/3):y\in L\}$ is a cover of $K.$
Let $\delta=\min \{r_y/3:y\in L\}.$
Now consider any $x$ whose distance from $K$ is at most $\delta.$ There exists $x'\in K$ with $d(x,x')< 3\delta /2$ and for this $x'$ there exists $y\in L$ such that $x'\in B_d(y,r_y/3).$ So $d(x',y)<r_y/3.$ So we have $$d(x,y)\le d(x,x')+d(x',y)\le$$ $$\le 3\delta /2+d(x',y)<$$ $$<3\delta /2+r_y/3\le$$ $$\le 3(r_y/3)/2+r_y/3<r_y.$$ So $x\in B_d(y,r_y)\subset G.$
Remark: If $K$ is non-empty and compact and the distance from $x$ to $K$ is at most $\delta$ then there exists $x'\in K$ with $d(x,x')\le \delta,$ but we do not need that in this Q.