Compact closures in the topology of compact convergence

Suppose first that \(\overline{\mathcal F}\) is compact in compact convergence. Evaluation at each \(x\) is continuous, so \(\{f(x):f\in\mathcal F\}\) has compact, hence bounded, closure. Thus \(\mathcal F\) is pointwise bounded. If equicontinuity failed at \(x\), there would be an \(\varepsilon>0\), a net \(f_i\in\mathcal F\), and a net \(x_i\to x\) with \(|f_i(x_i)-f_i(x)|\ge\varepsilon\). Let \(K\) be a compact neighborhood of \(x\); eventually \(x_i\in K\). Compactness of \(\overline{\mathcal F}\) gives a subnet \(f_i\to f\in\overline{\mathcal F}\) uniformly on \(K\). Then \[ |f_i(x_i)-f_i(x)|\le|f_i(x_i)-f(x_i)|+|f(x_i)-f(x)|+|f(x)-f_i(x)|\to0 \] along the subnet, by uniform convergence on \(K\) and continuity of \(f\), a contradiction. Hence \(\mathcal F\) is equicontinuous.

Conversely assume pointwise boundedness and equicontinuity. Embed \(C(X,\mathbb R^n)\) by restriction into the product \[ \prod_{K} C(K,\mathbb R^n), \] where \(K\) ranges over compact subsets and each factor has the uniform topology. For each \(K\), Ascoli says the restricted family has compact closure. Hence the closure of the image of \(\mathcal F\) lies in a product of compact spaces, so is compact by Tychonoff. The compatibility conditions on overlaps are closed. A compatible family defines a function on \(X\) whose restriction to every compact set is continuous; since every point of the locally compact Hausdorff space \(X\) has a compact neighborhood, that function is continuous. So the compatible families are exactly the restrictions of continuous functions, and restriction is a homeomorphism of \(C(X,\mathbb R^n)\), in the topology of compact convergence, onto them. Thus \(\overline{\mathcal F}\) is compact in compact convergence.

The two standard metrics on \(\mathbb R^n\) are uniformly equivalent, so pointwise boundedness and equicontinuity are unchanged by choosing either one.