= Solution
In the supremum-norm metric, $S$ is <totally bounded> if for every $\varepsilon>0$ finitely many norm balls of radius $\varepsilon$ cover it. It is uniformly bounded if some $M$ satisfies $|f(x)|\le M$ for all $f\in S,x\in K$. It is <equicontinuous> if for each $x\in K$ and $\varepsilon>0$ there is a neighbourhood $U$ of $x$ with $|f(y)-f(x)|<\varepsilon$ for all $f\in S,y\in U$. This definition works for compact Hausdorff $K$ without assuming it is metrizable.
Suppose $S$ is <totally bounded>. A finite radius-one net makes $\|f\|_\infty$ uniformly bounded by one plus the largest norm of a net centre. For equicontinuity, choose a finite $\varepsilon/3$ net. Continuity of each centre gives a common neighbourhood of $x$ on which all centre values vary by less than $\varepsilon/3$. Approximating any $f$ by its centre and using the <triangle inequality> at $x$ and $y$ bounds its variation by $\varepsilon$. Thus both conditions follow.
Conversely suppose those two conditions hold. For each $x$, choose a neighbourhood $U_x$ on which every function varies from its value at $x$ by less than $\varepsilon/3$. Compactness provides a finite cover $U_{x_1},\ldots,U_{x_m}$. The <vectors> $(f(x_1),\ldots,f(x_m))$ form a bounded subset of $\mathbb C^m$, so partition their possible values into finitely many boxes of diameter less than $\varepsilon/3$ in the maximum-coordinate norm. From each occupied box choose one function as a representative. If $f$ and its representative $h$ have sample values in the same box, and $y\in U_{x_j}$, then
$$
|f(y)-h(y)|\le|f(y)-f(x_j)|+|f(x_j)-h(x_j)|+|h(x_j)-h(y)|<\varepsilon.
$$
These finitely many representatives are a norm net, proving \b[total boundedness is equivalent to uniform boundedness and equicontinuity]. This supplies the finite-net part of the <Arzelà-Ascoli theorem> directly.
For a bounded set that is not <totally bounded>, take the standard unit <vectors> $S=\{e_n:n\ge1\}$ in the <Banach space> $\ell^2$. Their norms are one and their pairwise distances are $\sqrt2$. A ball of radius less than $\sqrt2/2$ contains at most one of them, so finitely many such balls cannot cover $S$.
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