Convergence of a sequence in compact convergence topology I have tried a proof for the following result from Munkres - General Topology and would like someone to go over it. There are some parts that I don't think are rigorous enough and would like help to write it better. Also i got stuck in the end.
Let $X$ be a topological space and $(Y,d)$ be a metric space and $Y^X$ be the set of all functions from $X$ to $Y$.
Deinition : Given an element $f$ of $Y^X$, a compact subspace $C$ of $X$ and a number $\epsilon>0$, let $B_C(f,\epsilon)$ denote the set of all those elements $g$ of $Y^X$ for which $$\text{sup }_{x\in C} \{d(f(x),g(x))\} < \epsilon$$ The sets $B_C(f,\epsilon)$ form a basis for a topology on $Y^X$ called the topology of compact convergence (TCC for short).
Theorem :  A sequence $f_n : X \rightarrow Y$ of functions converges to the function $f$ in the topology of compact convergence iff for each compact subspace $C$ of $X$, the sequence $f_n|_C$ converges uniformly to $f|_C$.
Proof : First consider a sequence of functions $f_n$ that converges to a function $f$ in the TCC. 
This means that given an $\epsilon >0$ I can find, for every compact set $K$ in $X$, an $n_K \in \mathbb{N}$ such that $f_n \in B_K(f,\epsilon)$ when ever $n \geq n_K$.
Let $C$ be a compact set in $X$. I need to prove that $f_n|_C \rightarrow f|_C$ uniformly. So given an $\epsilon>0$ I need to find an $n_0 \in \mathbb{N}$ such that $d(f_n(x),f(x)) < \epsilon$ whenevr $n \geq n_0$ for all $x \in C$. But now I can find an $n_C$ such that whenever $n\geq n_C$, $$f_n \in B_C(f, \epsilon) \Rightarrow \text{sup }_{x\in C }\{d(f_n(x),f(x))\} < \epsilon \Rightarrow d(f_n(x),f(x))<\epsilon$$ This $n_C$ depends only on $\epsilon$ and $C$ and not on each $x\in C$ (Should/Can there be a better way to say this? I thought that it is obvious). Thus the convergence is uniform. 
Conversely, suppose $f_n$ converges to $f$ uniformly on every compact $K$ in $X$.
Given an $\epsilon >0$, let $C$ be any compact set, then by the uniform convergence of $f_n$ we have that there exists a $n_C \in \mathbb{N}$ such that for all $x \in C$, $d(f_n(x),f(x)) <\epsilon$ whenever $n\geq n_C$. So $\text{sup }_{x\in C} \{d(f_n(x),f(x))\}\leq \epsilon$. But to say $f_n \in B_C(f,\epsilon)$ I need the last inequality to be strict so I am not sure how to conclude.
Thanks in advance.
 A: The first half of your argument is fine. You can make it just a little slicker with a minor adjustment:

Suppose first that $\langle f_n:n\in\Bbb N\rangle$ converges to $f$ in the TCC. Then for each $\epsilon>0$ and compact $K\subseteq X$ there is an $n_{K,\epsilon}\in\Bbb N$ such that $f_n\in B_K(f,\epsilon)$ whenever $n\ge n_{K,\epsilon}$. Now let $C\subseteq X$ be compact and $\epsilon>0$. Then if $n\ge n_{C,\epsilon}$, we have $f_n\in B_C(f,\epsilon)$, so $$d\big(f_n(x),f(x)\big)\le\sup_{y\in C}d\big(f_n(y),f(y)\big)<\epsilon$$ for each $y\in C$. $C$ is fixed here, so $n_{C,\epsilon}$ depends only on $\epsilon$, and it follows that $\langle f_n\upharpoonright C:n\in\Bbb N\rangle$ converges uniformly to $f\upharpoonright C$.

A very small (and rather standard) trick takes care of your difficulty with the second half. The uniform convergence of $\langle f_n\upharpoonright C:n\in\Bbb N\rangle$ to $f\upharpoonright C$ means that there is an $n_C\in\Bbb N$ such that $d\big(f_n(x),f(x)\big)<\color{blue}{\frac{\epsilon}2}$ whenever $x\in C$ and $n\ge n_C$, and you can now argue that
$$\sup_{x\in C}d\big(f_n(x),f(x)\big)\le\frac{\epsilon}2<\epsilon\;.$$
