 Solution

Suppose $C(K)$ is equicontinuous. At each $x\in K$, choose a neighbourhood $U$ such thatfor every $y\in U$ and every $f\in C(K)$. If some $y\in U$ differed from $x$, the Urysohn lemma for the compact Hausdorff space $K$ would give a continuous function with $f(x)=0$ and $f(y)=2$, a contradiction. Thus every singleton is open, so $K$ is discrete. A compact discrete space is finite, proving

Compactness is essential. Any infinite set with the discrete topology has equicontinuous $C(K)$, because the singleton neighbourhood $\{x\}$ works simultaneously for every function, but the space is infinite.