Arzelà–Ascoli $\implies$ Dini's theorem

Substracting the continuous limit we can assume that $f_n(x)$ are decreasing with limit $0$. This easily implies the equicontinuity of $\{f_n:n\in\mathbb N\}$. Indeed, given $x_0\in K$ and $\varepsilon>0$ there is $n\in\mathbb N$ with $0\le f_n(x_0)\le \varepsilon/4$. The continuity of $f_n$ then gives a neighbourhood $U$ of $x_0$ with $|f_n(x)-f_n(x_0)|\le\varepsilon/4$ for all $x\in U$ and thus $0\le f_n(x) = f_n(x)-f_n(x_0)+ f_n(x_0)\le \varepsilon/2$. For all $m\ge n$ and $x\in U$, this implies $$|f_m(x)-f_m(x_0)|\le f_m(x)+f_m(x_0)\le f_n(x)+f_n(x_0)\le \varepsilon.$$ Using the continuity of $f_1,\ldots,f_{n-1}$ we can finally make $U$ smaller to obtain the same continuity estimate for all $m\in\mathbb N$.

Let us now assume that $f_n$ does not converge uniformly to $0$, i.e., $\|f_{n(k)}\|_K\ge \delta$ for some $\delta>0$ and some strictly increasing sequence $n(k)$ of integers. Arzelá-Ascoli then implies that another subsequence $(f_{n(k(\ell))})_{\ell\in\mathbb N}$ converges uniformly to some continuous $g$ with $\|g\|_k\ge\delta$. But since uniform convergence implies pointwise convergence, $f_n\to 0$ pointwise, and pointwise limits are unique we obtain the contradiction $g=0$.

The argument is quite similar to a possible proof of the Arzelá-Ascoli theorem (which quite often is somewhat hidden behind technicalities): Compactness in the topology of pointwise convergence is due to Tychonov's theorem and equicontinuity implies that this topology coincides with the much finer topology of uniform convergence on the given set.