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

For each natural number $m\in\mathbb{N}$, let

where $B_{d_E}(0,m)$ is the open ball in the metric space $(\mathbb{R}^n,d_E)$.

By Open Ball in a Metric Space is Open, each $U_m$ 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 the family $(U_m)_{m\in\mathbb{N}}$ covers $\mathbb{R}^n$. Let $x\in\mathbb{R}^n$. The number $d_E(0,x)$ is a real number. By the Archimedean property, applied with $x=1$ and $y=d_E(0,x)$, there exists $m\in\mathbb{N}$ such that

Then $x\in U_m$. Thus

In particular, $(U_m)_{m\in\mathbb{N}}$ is an open cover of $A$ in $\mathbb{R}^n$.

Because $A$ is compact in $\mathbb{R}^n$, Compact Subset Criterion via Open Covers in the Ambient Space yields a natural number $k\in\mathbb{N}$ and indices $m_1,\dots,m_k\in\mathbb{N}$ such that

Set

If $y\in A$, then $y\in U_{m_j}$ for some $j\in\{1,\dots,k\}$, so

Hence

for every $y\in A$. Therefore $A$ is bounded in the metric space $(\mathbb{R}^n,d_E)$ by Bounded Subset of a Metric Space.