# Does "equalisers always closed" imply $T_2$?

Is there a non-$T_2$ space $(X,\tau)$ with the following property?

For all topological spaces $A$ and continous maps $f,g:A\to X$ the set $\{a\in A: f(a) = g(a)\} \subseteq A$ is closed.

Apply this to $A= X \times X$ with $f$ and $g$ the two projection on $X$ and it gives you that the diagonal $\Delta_X \subset X \times X$ is closed, and this is well known to be equivalent to $X$ is $T_2$.