Exercise 6c in section 50 Munkres' Topology textbook. The problem is as follows:
Given $f: X \to \mathbb{R}^N$ and given compact subspace $C$ of $X$ ($X$ is locally compact Hausdorff space with a countable basis); let:
$$U_\epsilon(C) = \{ f: \Delta(f|_C) < \epsilon \}$$
where $\Delta(f):= \sup\{ \operatorname{diam} f^{-1}(\{ z \}) : z\in f(X)\}$. (I assume that $f$ is continuous though it's not written explicitly).
Show that $U_\epsilon(C)$ is open in $\mathcal{C} (X,\mathbb{R}^N)$, the space of continuous functions from $X\to \mathbb{R}^N$ with the topology induced by the sup metric $\rho(f,g) = \sup \{ |f(x)-g(x)| : x\in X \}$. 
Now in the text there's a proof of the following for $X$ being compact metrizable space, Munkres proves that $U_\epsilon := \{ f: \Delta(f)<\epsilon\}$ is open in $\mathcal{C}(X,\mathbb{R}^N)$.
(it's on pages: 311-312).
My question: is the proof in the text the same as the proof for exercise 6c, if not then how to change it that it suits it?
 A: I will assume that we have some metric on the subspace $d$ which is compatible with the subspace topology induced from $X$. (The way I understand it, we need to have some metric on $X$ for the expression $\sup\{\operatorname{diam} f^{-1}(z)\}$ to make sense.)
At this point you already know this fact, which was shown as a part of the proof of Theorem 50.5.

If $(X,d)$ is a compact metric space then the set $$U_\varepsilon=\{f\in C(X,\mathbb R^n); \Delta(f)<\varepsilon\}$$ is open in the sup-metric.

This fact means that that if $f\in U_\varepsilon$, then also some $\delta$-ball around $f$ must be in $U_\varepsilon$. In the other words:

If $f\in C(X,\mathbb R^n)$ and $\Delta(f)<\varepsilon$, then there exists a real number $\delta>0$ such that $\Delta(g)<\varepsilon$ whenever $g\in C(X,\mathbb R^n)$ fulfills $$(\forall x\in X) d(f(x),g(x))<\delta.$$

Now we simply use this fact for the metric space $(C,d)$ to get a proof of the claim from Exercise 6c.
Proof. Let us assume that $f\in U_\varepsilon(C)$, i.e., $\Delta(f|_C)<\varepsilon$. Now there is a $\delta$-ball around $f|_C$ in the space $C(C,R^n)$ such that $\Delta(g)<\varepsilon$ for each $g\in C(C,R^n)$ which belongs to this ball.
Now if we take any $g\in C(X,R^n)$ such that 
$$(\forall x\in X) d(f(x),g(x))<\delta$$
then we also have 
$$(\forall x\in C) d(f(x),g(x))<\delta.$$
This means that the function $g|_C\in C(C,R^n)$ belongs to the $\delta$-ball with the center $f|_C$. And therefore $\Delta(g|_C)<\varepsilon$.
We have just shown that any function $f\in U_\varepsilon(C)$ has a $\delta$-ball around it which is subset of $U_\varepsilon(C)$. This means that $U_\varepsilon(C)$ is an open set. $\square$
