Proof of Compact Subset of $\mathbb{R}^n$ is Closed

To prove that $A$ is closed in $\mathbb{R}^n$, by Closed Subset of a Topological Space it is enough to show that the complement $\mathbb{R}^n\setminus A$ is open.

Let $x\in\mathbb{R}^n\setminus A$. For each point $a\in A$, the points $x$ and $a$ are distinct, so

Define

By Open Ball in a Metric Space is Open, each $U_a$ and $V_a$ is open in the metric space $(\mathbb{R}^n,d_E)$, hence open in the Euclidean sense by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$.

We claim that $U_a\cap V_a=\varnothing$ for every $a\in A$. Indeed, if $y\in U_a\cap V_a$, then by the triangle inequality from Euclidean Distance is a Metric on $\mathbb{R}^n$ we would have

which is impossible.

The family $(U_a)_{a\in A}$ covers $A$, because each point $a\in A$ belongs to $U_a$. Since $A$ is compact in $\mathbb{R}^n$, Compact Subset Criterion via Open Covers in the Ambient Space gives a natural number $k\in\mathbb{N}$ and points $a_1,\dots,a_k\in A$ such that

Set

Because the $V_{a_i}$ are open in the metric space $(\mathbb{R}^n,d_E)$ and metric open sets form a topology by Metric Open Sets Form a Topology, the set $V$ is open in the metric space $(\mathbb{R}^n,d_E)$. Therefore $V$ is open in the Euclidean sense by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$. Also $x\in V$.

We claim that $V\cap A=\varnothing$. If $y\in A$, then $y\in U_{a_i}$ for some $i\in\{1,\dots,k\}$. But $V\subseteq V_{a_i}$, so if $y\in V$ then $y\in U_{a_i}\cap V_{a_i}$, contradicting the disjointness proved above. Thus $V\cap A=\varnothing$.

Hence every point $x\in\mathbb{R}^n\setminus A$ has an open Euclidean neighborhood contained in $\mathbb{R}^n\setminus A$. Therefore $\mathbb{R}^n\setminus A$ is open, so $A$ is closed in $\mathbb{R}^n$.