On page 76, A guide to topology by Steven Krantz, there is a motivating question:
If $\mathscr E$ is a family of function from $S$ to $\mathbb R$, then under what circumstances is the mapping $$i: \mathscr E \times S \to \mathbb R$$ $$(f,s) \mapsto f(s)$$ continuous?
The answer is when $\mathscr E$ equipped with compact open topology.
So, let $f(s) \in U $ and $U$ is open in $\mathbb R$, we need to find $K$ and $W$ which are open in $\mathscr E$ and $S$ respectively such that $(f,s) \in K \times W$ and $i(K,W) \subseteq U$. I don't know how to proceed.
Below are screenshots:
Are there any additional conditions on $S$? If $S$ is locally compact and normal, for instance, it seems to work as follows: take any $(f,s)\in C(S,\mathbb{R})\times S$. Pick any neighbourhood of $f(s)$ in $\mathbb{R}$ in $S$ - say, $f(s)\in (a,b)$. Now if $f$ is continuous, we can pick $U$ - a neighbourhood of $s$ in $S$ with $f(U)\subset (a,b)$. By local compactness and normality choose an open $V\subset U$ with compact closure such that $s\in V\subset\overline{V}\subset U$, now take an open (in compact-open topology) neighbourhood $[\overline{V}, (a,b)]$ of $f$ and an open neighbourhood $V$ of $s$, and your $i$ is going to map $[\overline{V}, (a,b)]\times V$ into $(a,b)$, it seems. Can't get why it's true in general... Trying to build a counterexample ;).
EDIT Okay, here's an attempt for a counterexample...
Consider, for instance, the set $X = \mathbb{R}\cup \{x_0\}$, where $x_0\notin \mathbb{R}$, with the following topology (where only finite sets are compact): for any $x\in\mathbb{R}$ the set $\{x\}$ is open, and neighborhoods of $x_0$ are sets containing $x_0$ and having countable complements. This space is Hausdorff and not locally compact. Take the function $f_0:X\to \mathbb{R}$ which is zero at all points of $X$. Now take the point $0\in\mathbb{R}$ and consider its any neighbourhood $(a,b)$. It seems to me that $i$ won't map any neighborhood of $(f_0,x_0)$ into that neighborhood $(a,b)$ of $0 = f_0(x_0)$.
Indeed, take any finite set $\{y_1,y_2,...y_k, x_0\}$ (I included $x_0$ there because without loss of generality I can control of course the value of my functions at $x_0$ as well) and any neighborhood $U$ of $x_0$. Now take any point $x'\in U\backslash \{y_1,y_2,...y_k, x_0\}$. Define $f^*$ by $f^*(x') = b+1$ and $f^*(x) = 0$ for any $x\neq x'$. Then as far as I can see $f^*$ is continuous, and also $f^*$ belongs to the neighborhood $[\{y_1,y_2,...y_k, x_0\},(a,b)]$ of $f_0$. However, $i((f^*,x')) = f^*(x') = b+1 > b$ is not in $(a,b)$, while $(f^*,x')\in[\{y_1,y_2,...y_k, x_0\},(a,b)]\times U$. Are there any mistakes?
Here's yet another EDIT which deals with Brian M. Scott's useful comments. So I have to show in my example that for any finite collection $\{[F_i, (a_i,b_i)]\}_{i = 1}^n$, where $F_i = \{y^i_1,...,y^i_{k_i}\}$, $f_0\in \cap_{i=1}^n [F_i,(a_i,b_i)]$ and any open neighborhood $U$ of $x_0$ the function $i$ doesn't map $(\cap_{i=1}^n [F_i,(a_i,b_i)])\times U$ into $(a,b)$ (such sets form the base of the product topology on $C(X)\times X$, so that is clearly enough). In this case I can take my $f^*$ to be equal to $0$ everywhere except for some point $x'\in U\backslash (F_1\cup F_2\cup ... \cup F_n\cup \{x_0\})$, where it again equals, say, $b+1$.