= Solution
A family $\mathcal S\subseteq C(K)$ is <equicontinuous> if for every $x\in K$ and $\varepsilon>0$ there is a <neighborhood> $U$ of $x$ such that
$$
|f(y)-f(x)|<\varepsilon
$$
for every $y\in U$ and every $f\in\mathcal S$. The <Arzelà-Ascoli theorem> says that, for a <compact Hausdorff space> $K$, a subset of $C(K)$ has <compact closure> in the <uniform norm> if and only if it is equicontinuous and pointwise bounded. On compact $K$, equicontinuity and pointwise boundedness together imply <uniform boundedness>.
For necessity, a compact closure is <totally bounded>. Given $\varepsilon>0$, choose a finite $\varepsilon/3$-net $f_1,\ldots,f_m$. The finitely many <continuous functions> $f_j$ are simultaneously continuous near each point, and comparison with a nearby $f_j$ proves equicontinuity of the whole family. Evaluation at any fixed point is a continuous map $C(K)\to\mathbb R$ or $\mathbb C$, so compactness also gives pointwise boundedness.
Conversely, equicontinuity gives, around each $x\in K$, a neighborhood on which every $f\in\mathcal S$ oscillates by less than $\varepsilon/3$. <Compactness> supplies a finite such cover with selected points $x_1,\ldots,x_m$. Pointwise boundedness makes the set of vectors
$$
\{(f(x_1),\ldots,f(x_m)):f\in\mathcal S\}
$$
a bounded subset of a finite-dimensional <normed vector space>, so it has a finite $\varepsilon/3$-net. Choose one function of $\mathcal S$ for every nonempty cell of this net. If two functions have evaluation vectors in the same cell, then comparison at a selected $x_j$ and the two oscillation bounds show that their uniform distance is less than $\varepsilon$. Thus $\mathcal S$ is totally bounded. Since $C(K)$ is a <Banach space>, its closure is complete; a <complete totally bounded metric space is compact>, proving the theorem.
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