When does pointwise convergence imply uniform convergence?

The result is true as stated and is called Dini's Theorem. A proof can be found in Chapter III, Section 1.4 of these notes or indeed in this wikipedia article. (My notes are for a sophomore-junior level undergraduate class, so it is stated in the case in which $K$ is a closed interval. But it is clear that the argument works for any compact topological space.)

(For some reason this is one of those named theorems that tends to be assigned as an exercise or come up on exams...)

$\textbf{Dini's Theorem (with some extra conditions):}$ Let $K$ be a compact subset of $\mathbb{R}$ and suppose $f:K \to \mathbb{R}$ is continuous. Let $\{f_n\}$ be a monotonically decreasing sequence of continuous functions such that $f_n(x) \to f(x)$ pointwise. Prove that $f_n(x) \to f(x)$ uniformly.

$\textbf{Proof:}$ We want to show that $\forall \epsilon >0$ $\exists N$ such that $\forall n \ge N$ and $\forall x \in K$, $$|f_n(x) - f(x) | < \epsilon$$

Let $g_n(x) = f_n(x) - f(x)$. Then since $f_n(x) \searrow f(x)$ point wise we have that $g_n(x)$ is a monotonically decreasing sequence of continuous function. Now let $\epsilon > 0$ be fixed but arbitrary. We want to show the existence $N$ to satisfy the conditions of uniform convergence. Let, $$ E_n := \{ x \in K : g_n(x) = f_n(x) - f(x) < \epsilon \}$$ Then $E_n$ is open because it is the pre image of the continuous functions $g_n(x)$. Notice that $g_n(x)$ is continuous as it is the difference of continuous functions. We claim that $E_n$ is an ascending sequence of open sets. That is, $$ E_0 \subset E_1 \subset \dots \subset E_n \subset \dots$$ The sequence is ascending because $g_n(x)$ is decreasing. That is if $x \in E_n$, then $g_n(x) < \epsilon$, but this implies that $g_{n+1} (x) < \epsilon$, and so $x \in E_n$ implies that $x \in E_{n+1}$. Now as $n \to \infty$, we have that $g_n(x) \to 0$. Since $g_n(x) = 0 < \epsilon$, we have that the sets of $x \in K$ such that $g_n(x) < \epsilon$ will eventually be all of $K$. Thus, $$ \bigcup_{n=1}^{\infty} E_n \supset K$$ and so the above union is an open cover of $K$. Since $K$ is compact, there exists a finite subcover, such that, $$ \bigcup_{n=1}^N E_n \supset K$$ At this point, we should be happy as we have an appearance of $N$. Remember we are looking for $N$. $$K \subset \bigcup_{n=1}^N E_n$$ implies that $\forall x \in K$, we have that $|g_n(x)|= | f_n(x) - f(x)| < \epsilon$ Since $g_n(x)$ is decreasing, this immediately gives us that $\forall x \in K$, $\forall n > N$, $|f_n(x) - f(x)| < \epsilon$. Thus we have show $f_n(x)$ converge to $f(x)$ for arbitrary $\epsilon$ and so we are finished.

Your argument is very close.

Fix $\epsilon > 0$. Since $g_n(x) \downarrow 0$, there exists $m$ such that $g_m(x) \le \epsilon$. Now, for all $n_k \ge m$ we have $g_{n_k}(x_{n_k}) \le g_m(x_{n_k})$. As $k \to \infty$, the left side converges to $L$ by your construction. And by the continuity of $g_m$, the right side converges to $g_m(x) \le \epsilon$. So $L \le \epsilon$. Since $\epsilon$ was arbitrary, the proof is complete.

Edit: Incidentally, a uniformly convergent sequence is always equicontinuous, so equicontinuity does in fact hold. One could also prove it directly from the assumptions given (try it!).

Edit: As noted below by Monstrous Moonshine, your proof assumes that $K$ is sequentially compact, so it works for, say, metric spaces. But there is a gap if $K$ should be compact but not sequentially compact.