Topology on the set of Continuous Functions from a Compact to a Metric Space Let $X$ be a compact space, let $Y$ be a metric space, and let $F(X, Y)$ be the set
of all continuous functions from $X$ to $Y$. I want to show that the metric topology on $F(X, Y)$ induced by the metric $(f,g)=\sup_{x\in X}d(f(x),g(x))$ coincides with the compact open topology.
One inclusion is relatively easy but I'm having some trouble proving that all balls of the form $B_{d_{sup}}(f,\epsilon)$ for some $\epsilon>0$ and $f\in F(X,Y)$ are open with respect to the compact open topology. 
The subbasis elements of the compact open topology are of the form $F(K,O)\subset F(X,Y)$ for $K\subset X$ compact and $O\subset Y$ open. Is it correct saying this?
$$B_{d_{sup}}(f,\epsilon)=\bigcup_{x\in X}F(\{x\}, B(f(x),\epsilon))=F\left(X,\bigcup_{x\in X}B(f(x),\epsilon)\right)$$
If it's not, how can I prove this inclusion?
 A: Your statement concerning the subbasis elements is correct, but the equation
$$B_{d_{sup}}(f,\epsilon)=\bigcup_{x\in X}F(\{x\}, B(f(x),\epsilon))=F\left(X,\bigcup_{x\in X}B(f(x),\epsilon)\right)$$
is wrong. It is not true that $F(\{x\}, B(f(x),\epsilon)) \subset B_{d_{sup}}(f,\epsilon)$: In fact, $g \in F(\{x\}, B(f(x),\epsilon))$ means $g(x) \in B(f(x),\epsilon)$, i.e. $d(f(x),g(x)) < \epsilon$, but there is no reason why $d_{sup}(f,g) < \epsilon$.
In the following proof we have to assume that $X$ is compact Hausdorff.
So let $g \in B_{d_{sup}}(f,\epsilon)$, i.e. $\sup_{x \in X} d(f(x),g(x)) <\epsilon$. For each $x \in X$ there exists an open neighborhood $U_x$ of $x$ in $X$ such that $\max(d(f(x),f(y)),d(g(x),g(y)) < \delta = \frac{1}{2}(\epsilon -  \sup_{x \in X} d(f(x),g(x)))$ for all $y \in U_x$. Since $X$ is compact Hausdorff, it is locally compact and we find open neighborhoods $V_x$ of $x$ in $X$ such that $K_x = \overline {V_x}$ is compact and $K_x  \subset U_x$. There are finitely many $x_i \in X$ such that $X = \bigcup_i V_{x_i}$. The sets $W_i = F(K_i,B(f(x_i),\delta))$ are open in the compact-open topology. We have $g(K_i) \subset B(f(x_i),\delta)$: Given $y \in K_i$, we have $d(g(x_i),g(y)) < \delta$, thus
$$d(f(x_i),g(y)) \le d(f(x_i),g(x_i)) + d(g(x_i),g(y)) < \sup_{x \in X} d(f(x),g(x)) + \delta < \\ \sup_{x \in X} d(f(x),g(x)) + (\epsilon - \sup_{x \in X} d(f(x),g(x))  =\epsilon .$$ Therefore $g \in \bigcap_i W_i$.
Now let $h \in \bigcap_i W_i$. This means $h(K_i) \subset B(f(x_i),\delta)$ for all $i$. For $y \in K_i$ we therefore have $d(h(y),f(x_i)) < \delta$ and
$$d(f(y),h(y))  \le d(f(y),f(x_i)) + d(f(x_i),h(y)) < 2\delta < \epsilon .$$
Thus $h  \in \bigcap W_i  \subset B_{d_{sup}}(f,\epsilon)$.
