Read the whole proof: Munkres shows that all $y \in V$ obey $D(x,y) < \varepsilon$ (so that it lies in the ball and hence in $U$), and this computation really depends on the first $N$ coordinates of $y$, as the larger ones contribute so little to $D$ (due to the $\frac{1}{i}$ terms) that they can be ignored (and the sup reduces to a finite max).
Also, $y_1 \in (x_1-\varepsilon, x_1+\varepsilon)$, the first coordinate of $V$, ensures that $\bar{d}(y_1,x_1) < \varepsilon$ ; the diameter is $2\varepsilon$ but we only care about the distance to $\underline{x}$ we started with.
The sup metric that is also introduced in the book does not have this property that all large coordinates are irrelevant and Munkres also shows that the topology induced by that metric is actually not the product one. The $\frac{1}{i}$ terms are essential here, or we can also use a series of metrics that are weighted such that last terms are irrelevant as in this question.
You need to use that $C$, being compact in a Hausdorff space, is (a normal and hence) regular space, and in a regular space we have, whenever $x \in O$ open, an open $O'$ such that $x \in O' \subseteq \overline{O'} \subseteq O$.
Apply this to your $U \cap U'$ and inside we get the required $V$ almost for free from this regularity (open in open is open again, and $\overline{V}$ sits inside $C$ so is compact).
Added upon request:
Without using regularity directly, in Lemma 26.4 Munkres shows that whenever $Y$ is a compact subset of a Hausdorff space $X$ and $x_0 \notin Y$, there are open and disjoint $U$ and $V$ containing $x_0$ resp. $Y$, which is all the regularity we need:
We have $C$ compact and $U'$ the open neighbourhood of $x$ inside $C$. Then $C':=C \cap \left( X \setminus \left( U \cap U^\prime \right) \right)$ is compact (being closed in $C$) and $x \notin C'$. So by 26.4 we have open $x \in O_1$ and $O_2 \ \supseteq C'$ that are disjoint.
It's then not too hard to see that $V= U \cap U' \cap O_1$ is as required, it's certainly an open neighbourhood of $x$ and $\overline{V} \subseteq U$.
(For the last inclusion, which is clear (for me at least) from a picture, a more formal proof:
Suppose $y \in \overline{V}$, then in particular $y \in \overline{U'} \subseteq \overline{C}=C$ (using Hausdorff again!; also when we applied 26.4 of course). Also $y \in \overline{O_1}$, so $y \notin O_2$ (any point in $O_2$ has a neighbourhood, namely $O_2$, that is disjoint from $O_1$ to guarantee that it's not in the closure of $O_1$), and so $y \notin C'$ either. Knowing the definition of $C'$ and $y \in C$ we see that $y \notin X\setminus (U \cap U')$ which implies $y \in U' \cap U \subseteq U$ as required.
The regularity route is easier, so that's why I chose it first, but Munkres treats regularity after compactness, hence this addition.
Best Answer
As Brian M. Scott stated, the function $f: A \to \mathbb R$ sending $x$ to $d(x, Y \setminus V)$ is continuous on $A$ and $A$ is compact, so $f$ attains its minimum on $A$, at say $x \in A$. But if $d(x, Y \setminus V) = 0$, this means there exists a sequence of elements $\{y_n\}_{n \in \mathbb N}$ in $Y \setminus V$ satisfying $d(x,y_n) \to 0$. Since $Y \setminus V$ is closed, this implies $x \in Y \setminus V$, a contradiction since $A \subseteq V$. Therefore, we can pick $\varepsilon = \frac{d(x,Y \setminus V)}2$ to get an $\varepsilon$-neighborhood $U$ of $A$ contained in $V$. By definition, $d(U, Y \setminus V) = \frac{d(x, Y \setminus V)}2 > 0$, so $U$ is contained in $V$.
You won't be able to prove that $A \subsetneq U$ because it's not true in some cases. Take the example of $Y$ finite and discrete; $\varepsilon$-neighborhoods of $A$ are just $A$ for $\varepsilon$ small enough (smaller than any distance between any two distinct points, for example).
Hope that helps,