($\Rightarrow$: Compact implies closed and bounded.)

Closed: $\mathbb{R}^n$ is Hausdorff, and compact subsets of Hausdorff spaces are closed (Theorem 5.1(2)).

Bounded: The open cover $\{B(0, m)\}_{m=1}^{\infty}$ covers $K$. By compactness, finitely many suffice: $K \subseteq B(0, m_1) \cup \cdots \cup B(0, m_k) = B(0, M)$ where $M = \max(m_1, \ldots, m_k)$. So $K$ is bounded.

($\Leftarrow$: Closed and bounded implies compact.)

Step 1: We first prove that $[a, b]$ is compact for any $a \leq b$ in $\mathbb{R}$.

Let $\mathcal{U}$ be an open cover of $[a, b]$. Define: $s = \sup\{x \in [a, b] : [a, x] \text{ can be covered by finitely many sets in } \mathcal{U}\}.$

First, $s$ exists: the set is nonempty ($a$ is covered by some $U \in \mathcal{U}$, and $a$ plus a small interval lies in this $U$) and bounded by $b$.

Choose $U_0 \in \mathcal{U}$ with $s \in U_0$. Since $U_0$ is open, there exists $\epsilon > 0$ with $(s - \epsilon, s + \epsilon) \cap [a, b] \subseteq U_0$.

By definition of supremum, there exists $x \in (s - \epsilon, s]$ such that $[a, x]$ has a finite subcover $U_1, \ldots, U_k$ from $\mathcal{U}$. Then $[a, \min(s + \epsilon, b)] \subseteq U_1 \cup \cdots \cup U_k \cup U_0$, a finite subcover.

If $s < b$, this contradicts the definition of $s$ (we could extend further). So $s = b$, and $[a, b]$ has a finite subcover.

Step 2: By Tychonoff's theorem for finite products (Theorem 3.13), $[a_1, b_1] \times \cdots \times [a_n, b_n]$ is compact.

Step 3: Since $K$ is bounded, $K \subseteq [-M, M]^n$ for some $M > 0$. Since $[-M, M]^n$ is compact, and $K$ is a closed subset of a compact space, $K$ is compact.