Uniform convergence in the proof of properties of mollifier (Evan's approach) I am still trying to understand Evans' proof on the properties of mollifier.

In the proof of (iii), 

I understand that the crux of the proof is that uniform continuous function on compact set is uniform convergence. However, I do not understand why we need a middle space $W$, isn't that working with $V$ enough? What would happen if do not add such a $W$ in between $V$ and $U$? 
I was told that, the idea of having $W$ is that, if we can show convergence in $W$, then automatically we can show convergence in a smaller space. 
This is intuitive. But I am not convinced mathematically and I wish to see what would happen without $W$ in the proof.
 A: You are using that $f$ is uniform continuous when restricted to each open $W$ so that $W\subset\subset U$. 
Then for all $\delta >0$, there is $\epsilon_0 >0$ so that 
$$|f(x) - f(y)| <\delta$$
whenever $x, y\in W$ and $|x-y| < \epsilon_0$. Now if $x\in V$. Let $\epsilon_0$ be small so that $|y-x|<\epsilon_0 \Rightarrow y\in W$ (it is in general not true that $y\in V$). Then for all $\epsilon <\epsilon_0$,
$$\begin{split}
|f(x) - f^\epsilon(x)| &= \left| f(x) - \int_U f(y) \eta_\epsilon (y-x) dy\right| \\
&\le \int_U |f(x) - f(y)| \eta_\epsilon (y-x) dy \\
&\le \delta  
\end{split}$$
Note that in the second inequality we are using the uniform continuity of $f$ on $W$, not on $V$. As $x\in V$ is arbitrary, $f^\epsilon \to f$ uniformly on $V$.
Remark: The compactness of $V$ is used to guarantee the existence of such a $W$. Indeed, let $d>0$ be such that $d(v, \partial U) > d$ for all $v\in V$ (This $d$ can be found as $V$ is compact. Then let $W = V_{d/2} = \{ x\in U : d(x, V) < d/2\}$. Then $W$ is open and $\overline W$ is compact and is in $U$.   
