Let $X,Y$ be topological spaces, $C(X,Y)$ the set of continuous maps from $X$ to $Y$ with some topology $\mathcal{T}$. I'm trying to show that if the evaluation map $\operatorname{ev}:(f,x)\mapsto f(x)$ is continuous, then $\mathcal{T}$ contains the compact-open topology, with a subbasis given by sets of the form$$S(K,U):=\{f\in C(X,Y)\,|\,f(K)\subseteq U\}$$for $K\subseteq X$ compact and $U\subseteq Y$ open.
Here's what I've done so far:
Given some $f_0\in S(K,U)$, I'm trying to find an open neighbourhood of that point in $S(K,U)$. $\operatorname{ev}^{-1}(U)$ is given by$$\operatorname{ev}^{-1}(U)=\{(f,x)\in C(X,Y)\times X\,|\,f(x)\in U\}$$Assuming $S(K,U)$ is not empty, take some $f_0\in S(K,U)$ and some $x_0\in K$, clearly $(f_0,x_0)\in \operatorname{ev}^{-1}(U)$. Since $\operatorname{ev}^{-1}(U)$ is open in the product topology, there exist sets $V\subseteq C(X,Y)$ and $W\subseteq X$ open, such that $$V\times W\subseteq\operatorname{ev}^{-1}(U)$$If $K\subseteq W$, then $V\subseteq S(K,U)$, since by the condition above all $f\in W$ must map $W$ to $U$, so they must also map $K$ to $U$. So I would like to always find some open $W\subseteq X$ containing $K$ which has a partner $V$ such that $V\times W\subseteq\operatorname{ev}^{-1}(U)$ as above. I've tried taking $W=f_0^{-1}(U)$ which must contain $K$, but that has not seemed to help.
Any help is appreciated.