So here is proved that two continuous functions $f:X\to Y$ and $g:X\to Y$ from a space $X$ to an hausdorff space $Y$ are equal when they agree in a dense set $D$ of $X$. However it seem to me that the result could hold also when $Y$ is not hausdorff as the following argumentations seem to show.
So if $x\in X\setminus D$ then $x\in\operatorname{cl}D$ and so that it is the limit of a net $(x_\lambda)_{\lambda\in\Lambda}$ in $D$ and thus by continuity we conclude that $$ f(x)=\lim_{\lambda\in\Lambda}f(x_\lambda)=\lim_{\lambda\in\Lambda}g(x_\lambda)=g(x) $$ so that we finally conclude that $f$ and $g$ are equal.
So clearly in the above argumentations I do not use hausdorff-separability but only this relevant result which does not require that $Y$ is hausdorff: however I well know that if $Y$ is not Hausdorff then the limit of $f(x_\lambda)$ and $g(x_\lambda)$ are not generally unique and I realise that this could be a problem but unfortunately I was not able to find a counterexamle showing that the result could not hold when $Y$ is not an hausdoff space and thus I thought to put a specific queston. So could someone help me, please?
Consider $X=(\Bbb{R}, \tau_{std}) $ and $ Y=(\Bbb{R},\tau_{ind})$
$\tau_{ind}=\{\emptyset,\Bbb{R}\}$
Consider $f,g: X\to Y$ defined by
$f(x) =1 $ and $g(x) =\begin{cases}1 &x\in \Bbb{Q} \\0 & \text{otherwise}\end{cases}$
Then $f, g$ both are continuous ( as only open sets in $Y$ are $\emptyset$ and $\Bbb{R}$ )and $f=g $ on $\Bbb{Q}$ but $f\neq g$