= Solution
Suppose $C(K)$ is <equicontinuous>. At each $x\in K$, choose a neighbourhood $U$ such that
$$
|f(y)-f(x)|<1
$$
for 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
$$
\boxed{\ C(K)\text{ equicontinuous}\Longrightarrow K\text{ finite}.\ }
$$
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.
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