Proof of Heine-Borel Theorem in $\mathbb{R}^n$

We prove $(1)\Rightarrow(2)$ and $(2)\Rightarrow(1)$.

$(1)\Rightarrow(2)$ Assume that $A$ is compact in $\mathbb{R}^n$. Then Compact Subset of $\mathbb{R}^n$ is Bounded shows that $A$ is bounded as a subset of the metric space $(\mathbb{R}^n,d_E)$, and Compact Subset of $\mathbb{R}^n$ is Closed shows that $A$ is closed in $\mathbb{R}^n$. Hence condition $(2)$ holds.

$(2)\Rightarrow(1)$ Assume that $A$ is closed in $\mathbb{R}^n$ and bounded in the metric space $(\mathbb{R}^n,d_E)$. By Bounded Subset of a Metric Space, there exist a point $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and a real number $R>0$ such that

$$d_E(x,y)\le R$$

for every $y=(y_1,\dots,y_n)\in A$.

We show that $A$ is contained in a closed box. Let $y=(y_1,\dots,y_n)\in A$. Since

$$d_E(x,y)^2=\sum_{i=1}^n (y_i-x_i)^2\le R^2,$$

one has

$$(y_i-x_i)^2\le R^2$$

for every $i\in\{1,\dots,n\}$. Therefore

$$x_i-R\le y_i\le x_i+R$$

for every $i$. Hence

$$A\subseteq B,$$

where

$$B=\{z=(z_1,\dots,z_n)\in\mathbb{R}^n : x_i-R\le z_i\le x_i+R \text{ for every } i\in\{1,\dots,n\}\}.$$

The set $B$ is a closed box in $\mathbb{R}^n$, so it is compact in $\mathbb{R}^n$ by Closed Box in $\mathbb{R}^n$ is Compact. Since $A$ is closed in $\mathbb{R}^n$, it is also closed in the subspace $B$. Therefore Closed Subset of a Compact Space is Compact implies that $A$ is compact in $B$, hence compact in $\mathbb{R}^n$.

Thus conditions $(1)$ and $(2)$ are equivalent.