A compact subset of a metric space is closed and bounded

If $K = \emptyset$ then $K$ is bounded by the first clause of the definition, and it is closed because $X \setminus K = X$ is open.

Assume from now on that $K \ne \emptyset$ and fix $p \in K$; the family $(B(p,r))_{r \in P}$ indexed by the set $P$ of positive reals consists of open sets and covers $K$, since every $x \in K$ satisfies $d(p,x) < d(p,x) + 1$ and so lies in $B(p, d(p,x)+1)$.

The indexed characterisation gives $n \in \mathbb{N}$ and positive reals $r_0, \dots, r_n$ with $K \subseteq B(p,r_0) \cup \dots \cup B(p,r_n)$; putting $R := \max\{r_0, \dots, r_n\}$, a positive real, the balls with common centre $p$ are nested, so $K \subseteq B(p,R)$ and $K$ is bounded.

Boundedness being settled, take up closedness: let $q \in X \setminus K$ and for each $x \in K$ put $r_x := d(q,x)/2$, which is a positive real because $q \ne x$, and which satisfies $B(q,r_x) \cap B(x,r_x) = \emptyset$.

The family $(B(x,r_x))_{x \in K}$ consists of open subsets of $X$ and covers $K$, since $x \in B(x,r_x)$; so there are $n \in \mathbb{N}$ and $x_0, \dots, x_n \in K$ with $K \subseteq B(x_0, r_{x_0}) \cup \dots \cup B(x_n, r_{x_n})$.

Put $r := \min\{r_{x_0}, \dots, r_{x_n}\}$, a positive real.

Then $B(q,r) \cap K = \emptyset$: a point $y$ of the intersection would lie in $B(x_i, r_{x_i})$ for some $i \le n$ by step 4.1, and also in $B(q,r) \subseteq B(q, r_{x_i})$ by step 5.1, whereas those two balls are disjoint by step 3.1.

So every point of $X \setminus K$ has a ball around it inside $X \setminus K$, that set is open, and $K$ is closed; together with steps 1.1 and 2.1 this proves the theorem.