The title corresponds to Theorem 14.6 of General topology, by Stephen Willard.
Without loss of generality, we can assume that $f$ is onto too. Then, since $f$ is open, by a previous result it is enough to show that
$$A=\{(x_1,x_2): f(x_1)=f(x_2)\} $$
is closed in the product space $X\times X$.
The author picks $(x_1,x_2)\notin A$. Because of $f$ is closed, $\{f(x_2)\}$ is closed in $Y$ and hence $f^{-1}(f(x_2))$ is closed in $X$ by continuity.
Now we can invoke regularity and thus, there are disjoint open sets $U$ and $V$ such that $x\in U$ and $f^{-1}(f(x_2))\subseteq V$. And here is where I have my problem. Willard says: ''Since $f$ is closed, we can find a saturated open set in $X$ containing $f^{-1}(f(x_2))$ and contained in $V$;''.
How I can prove that? I have tried to use the fact that $f$ is both open and closed; then I have the relations
$f(\overline A)=\overline{f(A)}$
$f^{-1}(\overline A) = \overline{f^{-1}(A)}$
$f^{-1}(\operatorname{int}A)=\operatorname{int}f^{-1}(A)$,
but I don't know how to apply them in a satisfactory way. Since $X$ is regular, I can use that $V$ is closed if I want, so I have tried to apply the above to $V$ when it is open and also when it is closed, but I haven't found any solution.
Thanks in advance.
You can find the needed lemma and its proof (by me) here.
The regularity of $X$ gives us $U$ open with $x_1 \in U$ and open $V$ with $f^{-1}(\{f(x_2)\}) \subseteq V$ (as $f^{-1}(\{f(x_2)\})$ is a closed subset of $X$ that does not contain $x_1$), with $U \cap V = \emptyset$.
We apply the linked lemma with $p=f$ and $U=V$ to get an open subset $W$ of $Y$ with $f(x_2) \in W$ such that $f^{-1}(\{f(x_2)\}) \subset f^{-1}(W) \subset V$. (This $f^{-1}(W)$ is by definition a saturated open set, the one Willard mentions in his proof).
I hope this provides your missing link...