$\{ f_{n} \}$ is equicontinuous on $K$ $\implies$ for all $\epsilon > 0$, there exists $\delta > 0$ such that any $x, y \in K$ with $d(x, y) < \delta$ implies $\lvert f_{n}(x) - f_{n}(y) \rvert < \epsilon$ for all $n$.

For all $x \in K$, let $B_{x}$ denote the neighborhood of $x$ with radius $\delta$. Then $\{ B_{x} \}$ is an open cover of $K$. Since $K$ is compact, there exists a finite set of points $\{ p_{1}, \dots, p_{m} \}$ such that $\{ B_{p_{i}} \}$ is a finite subcover of $K$.

Now consider $B_{p_{i}}$. Let $M_{i}$ be an upper bound of $\{ \lvert f_{n}(p_{i}) \rvert \}$. Then for all $x \in B_{p_{i}}$, $|f_{n}(x)| < |f_{n}(p_{i})| + \epsilon \leq M_{i} + \epsilon$ for all $n$. Hence $M_{i} + \epsilon$ is a global upper bound of $\{ f_{n} \}$ on $B_{p_{i}}$.

Take $M = \max_{i} M_{i} + \epsilon < \infty$, then for all $x \in K$, $\lvert f_{n}(x) \rvert < M$ for all $n$. So $\{ f_{n} \}$ is uniformly bounded on $K$.

Choose any countable dense set $E$ on $K$. Then there exists a subsequence $\{ f_{n_{k}} \}$ of $\{ f_{n} \}$ such that $\{ f_{n_{k}} \}$ converges at every point of $E$. Let $\{ g_{k} \}$ denote $\{ f_{n_{k}} \}$.

Use the same $\{ B_{p_{i}} \}$ as in the first part. Pick any $x \in E$, then $x \in B_{p_{i}}$ for some $i$. Then $\{ g_{k}(x) \}$ converges, i.e., there exists a positive integer $N_{i}$ such that any $n, m > N_{i}$ means $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon / 3$. Now for any $y \in B_{p_{i}} \cap E$, we have that for any $n, m > N_{i}$, $$\lvert g_{n}(y) - g_{m}(y) \rvert = \lvert \left( g_{n}(y) - g_{n}(x) \right) + \left( g_{n}(x) - g_{m}(x) \right) + \left( g_{m}(x) - g_{m}(y) \right) \rvert < \epsilon$$ Now choose $N = \max_{i} N_{i} < \infty$. For any $x \in E$, $n, m > N$ implies $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon$. Hence $\{ g_{k} \}$ is uniformly convergent. Therefore, $\{ f_{n} \}$ contains a uniformly convergent subsequence.