Compact convergence and compact-open topologies on $\mathcal{C}(X,Y)$ coincide when $Y$ is metric [Proof Verification] As the title states, I'm currently working through the following exercise:

Let $(X,\tau)$ be a topological space and $(Y,d)$ a metric space. Then, the compact convergence and compact-open topologies on $\mathcal{C}(X,Y)$ coincide.

For completeness sake: the compact-open topology is generated by the sub-basis
$$
\{S(K,U) : K \subseteq X \text{ compact}, U \subseteq Y \text{open}\}
$$ 
with $S(K,U) = \{ f \in \mathcal{C}(X,Y) : f(K) \subseteq U\}$, and the compact convergence topology is generated by the basis of open balls restricted to each compact set, 
$$
\{B_K(f,\varepsilon) : \varepsilon > 0 , K \subseteq X \text{ compact}\}
$$
with $B_K(f,\varepsilon) = \{f \in \mathcal{C}(X,Y) : d(f(x),g(x)) < \varepsilon \ (\forall x \in K)\}$. What follows is my attempt. I would highly appreciate any comments on whether it is correct, and in that case, on how to make this argument more elegant.
Let's denote $\tau_{co}$ and $\tau_{c}$ for the compact-open and compact convergence topologies respectively and see both inclusions.


*

*($\tau_{co} \subseteq \tau_c$): let $S(K,U) \in \tau_{co}$. It suffices to see that $S(K,U)$ is open with respect to the compact convergence topology. Let $f \in S(K,U)$. Now, by defnition this means that $f(K) \subseteq U$ and so $f(K) \cap U^c = \emptyset$. Since $K$ is compact, $f(K)$ is compact and so it is closed in $Y$ (because it is metric). Hence we have that $\delta_f := d(f(K),U^c)$ is positive. This implies that $ B_K(f,\delta_f) \subseteq S(K,U)$. In effect, by the contrapositive if $g \in \mathcal{C}(X,Y)$ is such that $g(K) \cap U^c$ is non empty, then taking $g(x) \in g(K) \cap U^c$ we get
$$
d(f(x),g(x)) \geq d(f(K),U^c) = \delta_f.
$$
This proves that $B_K(f,\delta) \subseteq S(K,U)$ and so 
$$
S(K,U) = \bigcup_{f \in S(K,U)}B_K(f,\delta_f)
$$
is open for $\tau_c$.

*($\tau_{c} \subseteq \tau_{co}$): with the same idea, let $B_K(f,\varepsilon)$ be an open set for $\tau_{co}$ and let's see that for each $g \in B_K(f,\varepsilon)$, there exists $S(K',U') \in \tau_{co}$ with $g \in S(K',U') \subseteq B_K(f,\varepsilon)$. Since $K$ is compact and the mapping 
$$
K \hookrightarrow X \xrightarrow{1 \times 1} X \times X \xrightarrow{(f,g)} Y \times Y \xrightarrow{d} \mathbb{R}_{\geq 0}
$$ 
is continuous, it attains a maximum,
$$
\delta_g :=  \max_{x \in K}d(f(x),g(x)) < \varepsilon.
$$
Now, let $\mu = \varepsilon - \delta_g$, and note that the set
$$
U := \{y \in Y  : d(y,g(k)) < \mu \ , \ (\forall k \in K)\}
$$
is open: if $y\in U$ then $m_y := \max_{k \in K}d(y,g(k)) < \mu$. Thus, if $d(z,y) < \mu -m_y$,
$$
d(z,g(k)) \leq d(z,y) + d(y,g(k)) < \mu - m_y + m_y = \mu
$$
and therefore $B_{m_y}(y) \subseteq U$. Finally if $h \in S(K,U)$, then for each $x \in K$ we have that
$$
d(h(x),f(x)) \leq d(h(x),g(x)) + d(g(x),f(x)) < \mu + \delta_g = \varepsilon,
$$
which proves that $S(K,U) \subseteq B_K(f,\varepsilon)$.
Thoughts?
 A: $\newcommand{e}{\varepsilon}$
I will try to provide a proof to the second inclusion. I'm trying to adapt Willard's proof to this particular case of metric spaces as his proof uses the machinery of uniform spaces.
We want to construct an open set in the compact open topology such that it is included in $B_K (f,\e)$. For that consider a finite open cover of $f(K)$ given by  $B_{\e'}f(x_i)$ for $\e' < \dfrac{\e}{3}$. Now consider the open sets $U_i := \{ y : d(y,B_{\e'}(f(x_i))) < \epsilon' )\}$. Now with these constructions you can take the following compact and open sets for the c.o. topology. Let $K_i = K \cap f^{-1}(\overline{B_{\e'}(f(x_i))})$ (we take the closed ball so that $K_i$ is closed in a compact and so is compact). I say that we take as open set from the compact open topology the following set $\bigcap S(K_i,U_i)$. The idea is to control the bounds given by the triangular inequality.
Now it's time to check first that $f\in \bigcap S(K_i,U_i) $. For that we see that $f(K_i) \subset \overline{B_{\e'}(f(x_i))} $, so it's in $U_i$. 
Finally let's check that $\bigcap S(K_i,U_i) \subset B_K (f,\e) $. Let $g \in \bigcap S(K_i,U_i) $. As $K = \bigcup K_i$, if $x \in K$ then $x \in K_i$ for a certain $K_i$. As $f(x) \in \overline{B_{\e'}(f(x_i))}$ (we have seen this in the previous paragraph) and as $g(K_i) \subset U_i$ then $d(f(x_i),g(x)) \leq 2\e' < \e$ by definition of $U_i$. Now we use that $f(x) \in B_{\e'} f(x_i)$ to use the triangular inequality once more. So that $d(f(x),g(x)) \leq d(f(x_i),g(x)) + d(f(x_i),f(x)) \leq 3\e' < \e.$ 
