TITLE: Compact convergence topology coincide with compact-open topology if $Y$ is a metric space.
QUESTION [1 upvotes]: $\textbf{Theorem.}$ Let $X$ be a space and $(Y,d)$ be a metric space. On the set $C(X,Y)$, continuous functions from $X$ into $Y$, the compact convergence topology and the compact-open topology coincide.
I am stuck at a very simple step in the proof of the theorem above which I think many people will just assume.
Given $A$ a subset of $Y$ and $\epsilon > 0 $, then let $U(A,\epsilon)$ be the $\epsilon$ neighborhood of $A$. I am not sure how exactly $U(A,\epsilon)$ is defined as I couldn't find it in the book. But in my view it must be $\cup_{a \in A} B_{\epsilon}(a)$.
I want to show if $K$ is compact subset of $Y$ and $V$ is an open subset of $Y$ such that $K\subset V$. Then there exist an $\epsilon>0$ such that $U(K,\epsilon) \subset V$.
I know that the compact subsets of a metric space are closed and bounded. Hence, $K$ is closed and bounded. But since there are metric spaces in which a set can be both open and closed, and hence it is possible that $K=U$, then in that case $U(K,\epsilon) \not\subset V$ unless $U(K,\epsilon)=K$, which implies, $B_{\epsilon}(k) \subset K$ for all $k \in K$. 
If could help me prove the statement "If $K$ is compact subset of $Y$ and $V$ is an open subset of $Y$ such that $K\subset U$. Then there exist an $\epsilon>0$ such that $U(K,\epsilon) \subset V$."
I would really appreciate it.

REPLY [1 votes]: I am assuming you are using Topology by Munkres. The definition for the $\varepsilon$-neighborhood $U(K,\varepsilon)$ is defined on P. 177, Exercise 2(c), and the distance from a point $x$ to a set $A$ in a metric space is defined on P. 175.
The hints given by Dldier_ is adequate. I am just adding more details based on the proof given by Munkres on P. 285.
Consider the function $f:K\to\mathbb{R}$ defined by
$$f(x)=d(a,Y-V).$$
Since the domain $K$ is compact, by the extreme value theorem, there is some point $x\in K$ at which the function $f$ attains its minimum value $\varepsilon$.
In a metric space, for every point $x$ and subset $A$, it can be easily proven that
$$d(x,A)=0\iff x\in\bar{A}.$$
Note that $V$ is an open set, so its complement $Y-V$ is a closed set and hence $\overline{Y-V}=Y-V$. As we can see, we have $K\subseteq V$; that is, $K\cap(Y-V)=\emptyset$, so $f(x)>0$ for all $x\in K$. That is why $\varepsilon>0$.
Then we need to show that
$$U(K,\varepsilon)\subseteq V.$$
We can prove it by contradiction. Suppose $y\in U(K,\varepsilon)$ but $y\notin V$. Then for every $x\in K$, we have
$$d(x,y)\geq d(x,Y-V)\geq\varepsilon.$$
That is,
$$d(y,K)=\inf_{x\in K}{d(y,x)}\geq\varepsilon,$$
producing a contradiction.
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