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

We argue by induction on $n\in\mathbb{N}$.

For $n=1$, a closed box in $\mathbb{R}$ is exactly a closed interval $[a_1,b_1]$, which is compact by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$.

Assume now that the theorem is known in dimension $n-1$, and let

be a closed box in $\mathbb{R}^n$. Write

By the induction hypothesis, $C$ is compact in $\mathbb{R}^{n-1}$, and by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$ the interval $I$ is compact in $\mathbb{R}$. Hence, by Product of Two Compact Spaces is Compact, the product space $I\times C$ is compact when equipped with the product topology.

Define a map

Its image is exactly $B$.

We show that $f$ is continuous. Let $U\subseteq\mathbb{R}^n$ be open in the Euclidean sense, and let $(t,y)\in f^{-1}(U)$. Then $f(t,y)\in U$. By Euclidean Open Box Criterion in $\mathbb{R}^n$, there exists $\delta>0$ such that every point $z=(z_1,\dots,z_n)\in\mathbb{R}^n$ satisfying

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

Set

and

The set $(t-\delta,t+\delta)$ is open in $\mathbb{R}$, so $J$ is open in $I$ by the subspace topology. By the induction hypothesis together with Euclidean Open Box Criterion in $\mathbb{R}^n$, the set in braces is open in $\mathbb{R}^{n-1}$, so $V$ is open in $C$ by the subspace topology. Thus $J\times V$ is open in the product topology on $I\times C$ and contains $(t,y)$.

If $(s,w)\in J\times V$, then each coordinate of $f(s,w)$ differs from the corresponding coordinate of $f(t,y)$ by less than $\delta$, so by the choice of $\delta$ one has $f(s,w)\in U$. Therefore

which proves that $f^{-1}(U)$ is open. Hence $f$ is continuous.

Since $I\times C$ is compact and $f$ is continuous, Continuous Image of a Compact Space is Compact implies that its image $B$ is compact in $\mathbb{R}^n$. This completes the induction.