Let $(Y, \sigma)$ and $(X, \tau)$ be two topological spaces. If $X$ is regular and $f:X \to Y$ is a continuous, surjective, open and closed function, then $Y$ is Hausdorff.
I've already proved that if $f:W \to Z$ is an open and surjective function and the set $D=\{(x_1,x_2) |f(x_1)=f(x_2)\}$ is closed in $W \times W$, then $Z$ is Hausdorff.
Therefore I need only to prove that $D$ is closed in $X \times X$.
Because $X$ is normal I know that:
$\forall x \in X$, $\forall U \in \tau$ with $x \in U$, $\exists V \in \tau$ where $x \in V \subset \bar{V} \subset U$
$\forall B \subset X$, where $B$ is closed, $\forall x \in X\setminus B$ $\exists U, V \in \tau$ such that $x \in U$, $B \subset V$ and $U \cup V=\varnothing$.
I can't seem to find why either 1. or 2. in conjunction with any of the hypotheses of f imply that D is indeed closed. Any hint would be very much appreciated.