Open Set in $\mathbf{R}^n$ can be Covered by Countable Union of Compact Sets

I am aware of a couple answers, such as this one: Covering of open set by compact sets. My question is why exactly are the sets $$K_n := \{ x \in E: \operatorname{dist}(x, E^c) \geq \frac{1}{n}, \| x \| = d(0, x) \le n \}$$ closed with $E \subseteq \mathbf{R}^n$ being an open set? Is there a way to easily see this without using the sequential proof of a closed set (i.e. the limit of a sequence inside of a closed set stays inside)?

Let $f(x)=\operatorname{dist}\left(x,E^\complement\right)$ and let $g(x)=\|x\|$. Then $f$ and $g$ are continuous. Therefore, since $\left[\frac1n,\infty\right)$ and $[0,n]$ are closed sets, then so are the sets$$f^{-1}\left(\left[\frac1n,\infty\right)\right)=\left\{x\in E\,\middle|\,f(x)\geqslant\frac1n\right\}\tag1$$and$$g^{-1}\bigl([0,n]\bigr)=\left\{x\in E\,\middle|\,\right\|x\|\leqslant n\}.\tag2$$And your set is the intersection of $(1)$ and $(2)$.

A general topology approach that avoids the metric: $\Bbb R^n$ is locally compact Hausdorff. These properties both inherit to open subspaces $O$. Also, $\Bbb R^n$ is second countable and hence so is $O$. By Hausdorff local compactness of $O$, we can cover $O$ by open sets $U_x$ with $\overline{U_x} \subseteq O$ and such that the closure is compact. As $O$ is Lindelöf (implied by second countability), we can reduce these to a countable subcover and the corresponding closures are the required cover by countably many compact sets of $O$.