The more general version of this theorem in Munkres' 'Topology' (p. 290 – 2nd edition) states that
Given a locally compact Hausdorff space $X$ and a metric space $(Y,d)$; a family $\mathcal F$ of continuous functions has compact closure in $\mathcal C (X,Y)$ (topology of compact convergence) if and only if it is equicontinuous under $d$ and the sets
$$ \mathcal F _a = \{f(a) | f \in \mathcal F\} \qquad a \in X$$
have compact closure in $Y$.
Now I do not see why the Hausdorff condition on $X$ should be necessary? Why include it then? Am I maybe even missing something here (and there are counterexamples)?
btw if you are looking up the proof: Hausdorffness is needed for the evaluation map $e: X \times \mathcal C(X,Y) \to Y, \, e(x,f) = f(x)$ to be continuous. But the only thing really used in the proof is the continuity of $e_a: \mathcal C(X,Y) \to Y, \, e_a(f) = f(a)$ for fixed $a \in X$.
Cheers, S.L.
Best Answer
I think this question has been already been answered through the helpful comments. So thanks to Henno Brandsma and t.b.! This is just to finally tick it off.
My conclusion: It seems that $X$ being Hausdorff is rather a matter of convenience (maybe to avoid issues with the definition of local compactness for non-Hausdorff spaces, as pointed out in the comments), than a necessary condition.
Also this version of the theorem seems quite general enough for most uses.