My question is explicitly the following:
If $Y^{I}$ is given the compact-open topology is the map $Y^{I}\times I\to Y, (\gamma,t)\mapsto \gamma(t)$ continuous even if $Y$ is not Hausdorff?
In the case that $Y$ is Hausdorff one can quickly see that $Y^{I}$ is Hausdorff with the compact-open topology and the statement follows from the exponential law:
Exponential law
If $B$ is locally compact and $C$ is Hausdorff, then the map $$E:(A^{B})^{C}\to A^{B\times C}, \qquad f\mapsto \big(\ (b,c)\mapsto f(c)\,(b)\ \big)$$ is well defined and a homeomorphism, where the spaces of continuous maps have been given the compact-open topology.
The identity function $Y^{I}\to Y^{I}$ is continuous and the above map associates it to the function $Y^{I}\times I\to Y, (\gamma,t)\mapsto \mathrm{id}(\gamma)\,(t)=\gamma(t)$, so the evaluation is continuous.
Some context: The evaluation at $t$ for a fixed $t$ is continuous as a map $Y^{I}\to Y$ regardless of whether or not $Y$ is Hausdorff. This was an exercise and I wonder if this generalisation is always true.