Suppose $K$ is compact. Consider the open cover

$\mathcal{U} = \{(-n, n) \mid n \in \mathbb{N}\}.$

Each $(-n, n)$ is an open interval, and $\bigcup_{n=1}^\infty (-n, n) = \mathbb{R} \supseteq K$, so $\mathcal{U}$ is an open cover of $K$.

By compactness, there exist finitely many $n_1, \ldots, n_k \in \mathbb{N}$ such that

$K \subseteq (-n_1, n_1) \cup \cdots \cup (-n_k, n_k) = (-N, N),$

where $N = \max(n_1, \ldots, n_k)$. Thus $K$ is bounded.

We show $\mathbb{R} \setminus K$ is open. Let $x \in \mathbb{R} \setminus K$. For each $y \in K$, define

$U_y = \left(-\infty, \frac{x + y}{2}\right) \quad \text{and} \quad V_y = \left(\frac{x + y}{2}, \infty\right).$

Then $U_y$ and $V_y$ are open, $y \in U_y$, and $x \in V_y$. Note that $U_y \cap V_y = \varnothing$.

The collection $\{U_y \mid y \in K\}$ is an open cover of $K$. By compactness, finitely many cover $K$:

$K \subseteq U_{y_1} \cup \cdots \cup U_{y_m}.$

Define $V = V_{y_1} \cap \cdots \cap V_{y_m}$ (a finite intersection of open sets, hence open). Then $x \in V$ (since $x \in V_{y_i}$ for all $i$).

Claim: $V \cap K = \varnothing$.

If $z \in V \cap K$, then $z \in V$ and $z \in K$. Since $K \subseteq U_{y_1} \cup \cdots \cup U_{y_m}$, we have $z \in U_{y_j}$ for some $j$. But $V \subseteq V_{y_j}$, so $z \in V_{y_j}$. This contradicts $U_{y_j} \cap V_{y_j} = \varnothing$.

Thus $V$ is an open neighborhood of $x$ contained in $\mathbb{R} \setminus K$. Hence $\mathbb{R} \setminus K$ is open, so $K$ is closed.

Suppose $K$ is closed and bounded. Since $K$ is bounded, $K \subseteq [a, b]$ for some closed interval $[a, b]$.

Lemma: $[a, b]$ is compact.

Proof of Lemma: Let $\{U_\alpha\}_{\alpha \in I}$ be an open cover of $[a, b]$. Define

$S = \{x \in [a, b] \mid [a, x] \text{ can be covered by finitely many } U_\alpha\}.$

We show $b \in S$, which proves the lemma.

Step 1: $S \neq \varnothing$. Since $a \in [a, b]$, there exists $\alpha_0$ such that $a \in U_{\alpha_0}$. Since $U_{\alpha_0}$ is open, there exists $\epsilon > 0$ such that $(a - \epsilon, a + \epsilon) \subseteq U_{\alpha_0}$. Thus $[a, \min(a + \epsilon/2, b)] \subseteq U_{\alpha_0}$, so $\min(a + \epsilon/2, b) \in S$. Hence $S \neq \varnothing$.

Step 2: $S$ is bounded above (by $b$), so $c = \sup S$ exists by completeness.

Step 3: $c \in S$. Since $c \in [a, b]$, there exists $\alpha_1$ such that $c \in U_{\alpha_1}$. Since $U_{\alpha_1}$ is open, there exists $\delta > 0$ such that $(c - \delta, c + \delta) \subseteq U_{\alpha_1}$.

Since $c = \sup S$, there exists $x \in S$ with $c - \delta < x \leq c$. By definition of $S$, $[a, x]$ can be covered by finitely many $U_\alpha$, say $U_{\alpha_2}, \ldots, U_{\alpha_n}$. Then

$[a, c] \subseteq [a, x] \cup (c - \delta, c + \delta) \subseteq U_{\alpha_2} \cup \cdots \cup U_{\alpha_n} \cup U_{\alpha_1}.$

Thus $c \in S$.

Step 4: $c = b$. Suppose $c < b$. By Step 3, $[a, c]$ can be covered by finitely many $U_\alpha$. Since $c \in U_{\alpha_1}$ and $U_{\alpha_1}$ is open with $(c - \delta, c + \delta) \subseteq U_{\alpha_1}$, we have

$[a, \min(c + \delta/2, b)] \subseteq [a, c] \cup (c, c + \delta/2) \subseteq U_{\alpha_2} \cup \cdots \cup U_{\alpha_n} \cup U_{\alpha_1}.$

Thus $\min(c + \delta/2, b) \in S$. Since $c < \min(c + \delta/2, b)$, this contradicts $c = \sup S$. Hence $c = b$.

Conclusion: $b \in S$, so $[a, b]$ can be covered by finitely many $U_\alpha$. Thus $[a, b]$ is compact. $\square$

Back to the main proof: We've shown $[a, b]$ is compact. Since $K \subseteq [a, b]$ and $K$ is closed, $K$ is a closed subset of a compact set, hence compact.

(Fact: Closed subsets of compact sets are compact. Proof: If $K \subseteq [a, b]$ is closed and $\{U_\alpha\}$ is an open cover of $K$, then $\{U_\alpha\} \cup \{\mathbb{R} \setminus K\}$ is an open cover of $[a, b]$. Extract a finite subcover; removing $\mathbb{R} \setminus K$ if necessary gives a finite subcover of $K$.)