= Solution
Assume first that $\overline{\mathcal F}$ is compact in the <supremum norm>. A finite norm-net shows that the norms of its members are bounded, so pointwise boundedness follows. Given $x$ and $\varepsilon>0$, choose a finite $\varepsilon/3$-net $g_1,\ldots,g_N$ in the closure. Their continuity gives a common neighborhood $U$ of $x$ where $|g_j(y)-g_j(x)|<\varepsilon/3$ for every $j$. For any $f\in\mathcal F$, choose $g_j$ with $\|f-g_j\|_\infty<\varepsilon/3$; the <triangle inequality> then gives $|f(y)-f(x)|<\varepsilon$. Thus the family is equicontinuous.
Conversely, suppose pointwise boundedness and equicontinuity. For a prescribed $\varepsilon>0$, choose neighborhoods $U_x$ where every $f$ varies from $f(x)$ by less than $\varepsilon/3$. Compactness supplies a finite subcover $U_{x_1},\ldots,U_{x_N}$. Pointwise boundedness makes the set of vectors $(f(x_1),\ldots,f(x_N))$ bounded in $\mathbb R^N$. Partition a bounding box into finitely many boxes of coordinate diameter less than $\varepsilon/3$, and choose one function from each occupied box. If $f,g$ belong to the same box and $y\in U_{x_j}$, then
$$
|f(y)-g(y)|\le |f(y)-f(x_j)|+|f(x_j)-g(x_j)|+|g(x_j)-g(y)|<\varepsilon.
$$
The chosen functions are therefore a finite $\varepsilon$-net in the <supremum norm>. Thus $\mathcal F$ is totally bounded. Its closure is complete and totally bounded, hence compact. To recall the last metric-space fact, successively selecting subsequences in balls of radii $2^{-j}$ produces a Cauchy diagonal subsequence from every sequence; completeness supplies its limit. This proves both directions of the <Arzelà-Ascoli theorem>.
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