Counter example with compact open convergence I would like an (easy?) example (if any) of a topological space $X$, a metric space $Y$, a sequence of funcions $f_n:X\to Y$, and a function $f:X\to Y$  such that:
1) each $f_n$ is continuous
2) $f_n\to f$ with respect to the compact open topology
3) $f$ is not continuous.
In other words, I search for a counter example to the sentence "compact open limit of continuous functions is continuous", which, as far as I can tell, it is not always true.
 A: If $X$ is a $k$-space (aka compactly generated), then $f$ is continuous on $X$ iff $f\restriction_K$ is continuous on $K$ for all compact subsets of $X$. And for compact domain $K$, the compact-open topology is generated by the sup-metric and it's well-known that there the uniform limit of continuous functions is continuous. So your positive result will hold on all $k$-spaces (which includes metric and first-countable spaces, locally compact Hausdorff spaces, as its most typical subclasses.) 
I think $X$ in the co-countable topology (not compactly generated) can be used to build a counterexample.
A: I would like to discuss the following example:
$X=\{0,1,1/2,1/3,1/4...\}$ is a compact metric space (or any converging sequence together with its limit).
$Y=\{0,1\}$. It is clearly metric.
Now set $f_n=\mathcal 1_{[0,1/n]}$ the restriction to $X$ of the characteristic function of $[0,1]$, and $f=1_{\{0\}}$ be the function that holds $1$ in $0$ and zero elsewhere.
$||f_n-f||_{\sup}=1$ because $f_n(1/n)=1$ and $f(1/n)=0$.
On the other hand, $f_n$ seems to me converging to $f$ in the compact-open topology. This would provide a counterexample becauuse $f_n$ are continuous and $f$ is not.
Proof that $f_n\to f$ in the compact-open topology:
1) Non trival open sets of $Y$ are $\{0\}$ and $\{1\}$,
2) Compacta of $X$ are: 
2.1) finite subsets, or
2.2)  sets of the form $K=\{x_n\}\cup\{0\}$ where $x_n\to 0$ in $X$. 
Therefore the non-trivial open neighborhoods of $f$ in the compact open topology are of the form
$V=\{g: g(0)=1,g(x_1)=....=g(x_k)=0\}$ for some $x_1,\dots,x_k\in X$ different from zero or
$V=\{g: g(x_1)=....=g(x_k)=0\}$ for some $x_1,\dots,x_k\in X$ different from zero.
Given such an open sets $f_n\in V$ eventually on $n$, precisely from $n$ bigger than $\frac{1}{\min(x_1,...,x_k)}$
