It seems to me that you can w.l.o.g. assume that $U$ is open and all the $V_i$ are open.
In other words (making this wlog a bit more formal) you can just take the interiors $\mathring{U}$ and $\mathring{V}_i$ and solve the problem with the interiors.
Once you have solved the problem with the $\mathring{U}$ and the $\mathring{V}_i$ instead of $U$ and $V$ you have found functions $\zeta_i \in C^{\infty}$ such that
$$
\begin{cases}
0 \leq \zeta_i \leq 1\\
\text{supp}(\zeta_i) \subset \mathring{V}_i\\
\sum_{i} \zeta_i = 1 \text{ on } \mathring{U}.
\end{cases}
$$
Now you can use continuity of the $\zeta_i$ and $\sum_{i} \zeta_i$ to conclude that you actually solved the problem for the original $U$ and $V_i$.
Solving the problem for the interiors can be done in a few steps (I think!! but it has been some time since I did this proof, but it can be found online probably, but some hints might be helpful if you want figure it out without full solutions first):
- First conclude that $U \subset \subset \cup_i V_i \quad$ also holds when you substitute $U$ and $V_i$ with the interiors.
- Use problem 5 in your image to first find $\xi_i$ for every $V_i$ (actually the interior), this can be done because you are solving the problem with the interiors (which are of course open!)
- Some tricks to define $\zeta_i$ using a smart division of the form $\zeta_i := \frac{\xi_{i}}{\sum_{j} \xi_j}$.
Edit:
An update after your comment.
Indeed, we have $\cup_i(\text{int}(V_i)) \subset \text{int}(\cup_i(V_i))$, but in general not the other way around, so your comment is valid and I did not think of this beforehand.
Luckily however, things work out in the following way:
Suppose we have a point $x \in \text{int}(\cup_i(V_i)) - \cup_i(\text{int}(V_i))$ and $x \in \overline{U}$.
This means that $x \not \in \text{int}V_i$ for all $i \in \{1,\dots,n\}$ and $x \in \partial V_i$ for at least one $i$.
However, in this case a partition of unity subordinate to $\{V_i\}_i$ cannot exist: Indeed for arbitrary $i \in \{1,\dots,n\}$ we have $\zeta_i$ with support in $V_i$ which means that $\zeta_i(x) = 0$, since either $x \in \partial V_i$ or $x \not \in V_i$.
This implies we have $\sum_i \zeta_i(x) = 0.$
However, since $x \in \overline{U}$ we get by continuity $\sum_i \zeta_i(x) = 1$, a contradiction.
So we are not interested in this case, since no interesting partition of unity exists, so we can wlog assume that $x \in \overline{U} \implies x \in int(V_i)$ for some $i$, and I believe this solves the problem you correctly stated in your comment.
The key fact that he's using without clearly stating it is that a uniformly continuous function on a dense subset of a metric space uniquely extends to a uniformly continuous function on the full metric space. If Evans assumed that $u$ were uniformly continuous on all of $U$, then the density of $U$ in $\bar{U}$ would give us the extension of $u$ to $\partial U$ immediately. However, he only assumes the local uniform continuity condition, so we need to do a bit more work. The idea is that if we fix a point $p \in \partial U$ and consider the ball $B(p,r)$, then $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$, and our function $u$ is uniformly continuous in the former set by assumption. So, we can extend it uniquely to a continuous function on $\bar{U} \cap B(p,r)$, which in particular allows us to uniquely assign values to $u$ on $\partial U \cap B(p,r)$. This works for all $p \in \partial U$, and so we can uniquely extend $u$ to $\partial U$ using this trickery, and the resulting extension is continuous.
On the other hand, you can ask about the space
$$
\dot{C}(\bar{U}) = \{u : \bar{U} \to \mathbb{R} \;\vert\; u \text{ is continuous on } \bar{U}\}
$$
(I've added the dot to distinguish this from his space). We've just seen that any $u \in C(\bar{U})$ (his space) uniquely extends to define an element of $\dot{C}(\bar{U})$. On the other hand, any $u \in \dot{C}(\bar{U})$ restricted to $U$ defines an element of $C(\bar{U})$ since $u$ will be uniformly continuous on compact subsets of $\bar{U}$, and any bounded subset of $U$ will be contained in such a compact subset.
Ultimately, what we've done is built a linear extension isomorphism (you can check easily that it's linear) $E : C(\bar{U}) \to \dot{C}(\bar{U})$, and a linear restriction isomorphism $R: \dot{C}(\bar{U}) \to C(\bar{U})$ such that $ER = I$ and $RE = I$. In other words, these spaces are naturally isomorphic via these extension and restriction processes. So, what's the point? In some sense the space $C(\bar{U})$ is preferable because it only requires the functions involved to be defined on $U$. This means that statements like $W^{k,p}(U) \hookrightarrow C(\bar{U})$ (for appropriate Sobolev parameters $k,p$) are slightly more natural because the functions in the Sobolev space are only defined on $U$ a priori. If we were to write $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$, then there would be a slight abuse of notation in that the functions in the first space are only defined a priori in the smaller space $U$. This isn't a big deal, as we've seen above with $E$, and many texts simply gloss over this issue and take the statement $W^{k,p}(U) \hookrightarrow \dot{C}(\bar{U})$ to mean
$$
W^{k,p}(U) \hookrightarrow C(\bar{U}) \xrightarrow{E} \dot{C}(\bar{U}).
$$
EDIT: Here is a proof of the claim that $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$ in any metric space. Let $\varepsilon >0$ and pick a point $y \in \bar{U} \cap B(p,r)$. We know that $\bar{U} = U \cup \partial U$. If $y \in U \cap B(p,r)$, then we simply pick $x = y \in U \cap B(p,r)$ and note that $d(x,y) = 0 < \varepsilon$. Suppose, then, that $y \in \partial U \cap B(p,r)$. By definition of the boundary, we can pick $x \in U \cap B(y,s)$ for $s = \min\{\varepsilon, r-d(y,p) \}$. Then, certainly, $d(x,y) < s \le \varepsilon$, but also $d(x,p) \le d(x,y) + d(y,p) < s + d(y,p) \le r$, so $x \in U \cap B(p,r)$. We have now shown that for every $y \in \bar{U} \cap B(p,r)$ and $\varepsilon >0$ there exists $x \in U \cap B(p,r)$ such that $d(x,y) < \varepsilon$, so $U \cap B(p,r)$ is dense in $\bar{U} \cap B(p,r)$.
Best Answer
From the previous theorem, we know how to approximate $u$ by mollification; but mollification requires some room (I like to think of it as having a smudging effect), so we need to be away from the boundary. To approximate $u$ on the entirety of $U$, we would like to divide $U$ into infinitely many open sets separated from its boundary, mollify $u$ on each set and add these mollifications to obtain an approximation of $u$. A partition of unity subordinate to the covering of $U$ by such open sets is the standard tool to achieve this.
We must then ask when a partition of unity subordinate to a given cover of $U$ exists. A sufficient condition is that the cover be locally finite (see this), meaning that every point of $U$ has a neighborhood intersecting only finitely many members of the cover. Clearly, $\{U_i\}$ is not locally finite and will not even admit a locally finite refinement. By contrast, $\{V_i\}$ is a locally finite cover, and hence admits a partition of unity satisfying the desired properties.
The $V_i's$ are not meant to be disjoint, for if $U$ is connected, then it cannot be the union of disjoint open sets. This explains why $V_i = U_{i+3} - \overline{U}_{i+1}$. I hope this addresses 1.
As for 2., your choice of $V_0$ indeed meets the requirements.
I addressed 3 somewhat above, but partitions of unity are a standard construction in topology and we can create them by Urysohn's lemma and mollification.