Help with exercise from Munkres's of general version of Arzela's Theorem So i have been trying to do an exercise from Munkres's that goes like this:
Let $X$ be an Hausdorff space that is $\sigma$-compact; Let $f_n$ be a sequence of functions $f_n : X \rightarrow \mathbb{R}^k$. If the collection is pointwise bounded and equicontinuous, then the sequence $f_n$ has a subsequence that converges, in the topology of compact convergence, to a continuous function. And he gives a hint that is to prove that $C(X,\mathbb{R}^k)$ is first-countable.
So with this said assuming that $C(X,\mathbb{R}^k)$ is first-countable this was easy to prove because compact is going to implie sequentially compact in the family of equicontinuous functions , etc ...
My problem is that i cant quite seem to prove that $C(X,\mathbb{R}^k)$ is first countable. So we know that $X$ is $\sigma$-compact so we can assume that $X= \cup^{n=1}_\infty K_n$ , with $K_n$ compact for any $n$ and $K_n \subset K_{n+1}$. So my idea was that for a fixed continuous function the basis at that point would be the sets of the form $B_{K_n}(f,1/n)$. With this my idea would be try to prove that for any compact set $K$ then there exists an $n$ such that $K \subset K_n$ and so $B_{K_n}(f,1/n^*) \subset B_K(f,\epsilon)$ for some $n^*$. Now im not very familiarized with $\sigma$-compact spaces but this doenst seem to be trivially true, that any compact set is inside some $K_n$, so if anyone knows any counterexample it would also be aprecciated, if it were true the problem would be easy and i dont understand why we would need the condition for the space to be hausdorff and the fact that we are restricting ourselves to $C(X,\mathbb{R}^k)$. So any tips on this step of the proof would be aprecciated, i just wanted something that would get my head rolling and start to understand why we need these conditions. Thanks in advance!
 A: Munkres defines $\sigma$-compact on page 289 (2nd edition), in exercise 10: $X$ is a countable union of open sets $O_n$ such that $O_n \subseteq K_n$ with $K_n$ compact. This notion is in other books often called hemi-compact (again Munkres chooses non-standard terminology, as he is wont to do)
This is quite a bit stronger than just being a countable union of compact sets: $\Bbb Q$ is that trivially, but is not $\sigma$-compact in the strong sense (i.e. hemicompact) as no open set is contained in a compact subset. Clearly, $X$ hemicompact implies that $X$ is locally compact.
For $X$ hemicompact, we do have that every compact $K$ is contained in some $K_n$ when we choose the $O_n$ and $K_n$ increasingly as we can do WLOG. This because the $O_n$ must then have a finite (and thus singleton) subcover, etc.
And for $X$ hemicompact we can indeed show that $Y^X$ for $Y$ metric is first countable (even metrisable, as this mentioned exercise 10 asks you to show) as the $B_{K_n}(f, \frac{1}{m})$ then form a countable local base at $f$. 
