Theorem 46.8 Munkres Topology In the proof of Theorem 46.8 Munkres Topology following is claimed (not exact copy but exact interpretation)

If A is any compact subset of a metric space Y and V is any  open subset of Y containing A then there is a $\epsilon>0 $ such that the $\epsilon-$neighborhood of A is contained in V.

The mentioned claim is true provided $d(A, X-V) >0$ is positive. 1- But how to prove that? And 2- how any open set containing a compact set is a proper superset?
 A: 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,
A: A direct proof which does not relyy on the continuity of the distance function (a handy fact though that is..):
If $A \subseteq V$ for each $a \in A$ find $r_a>0$ such that $B(a,2r_a) \subseteq V$ (by openness of $V$ and the fact that $a \in A$ implies $a \in V$).
By compactness of $A$ there are finitely many $a_1, \ldots a_n \in A$ so that
$$A \subseteq \bigcup_{i=1}^n B(a_i, r_{a_i})$$
and define $r=\min_{i=1}^n r_{a_i} >0$.
If now $x$ in $U(A,r)$ there is some $a \in A$ such that $d(x,a) < r$. This $a$ lies in some $B(a_i, r_{a_i})$ so that $d(x,r_{a_i}) \le d(x,a) + d(a,a_i) < r + r_{a_i} < 2r_{a_i}$ and so $x \in B(a_i, 2r_{a_i}) \subseteq V$. This shows that $U(a,r) \subseteq V$ as required.
