I'm looking for counterexamples such that
$f: X \rightarrow Y$ is a continuous closed surjective map with $X$ Hausdorff and $Y$ non-Hausdorff
I know perfect maps preserve Hausdorff property. The only extra assumption that is crucial in the proof is $f^{-1}\{y\}$ is compact for all $y\in Y$. That's why I try to drop this assumption. The closest question I can find is Is Hausdorffness preserved under continuous surjective open mappings?, which almost excites me except for the open. I looked at the answers in it and tried them one by one. Unfortunately, none of them work or can be mimicked. I also tried the quotient map with $Y$ being the "line with infinitely many origins". Also failed. Any hint would be appreciated.
Let $X$ be a Tychonoff ($T_{3\frac12}$) space that is not normal (like the Sorgenfrey plane $\mathbb{S}^2$, e.g.), and let $A$ and $B$ be two disjoint closed sets that cannot be separated by open sets.
Define an equivalence relation $R$ on $X$ by specifying its classes: $A$, $B$, and $\{x\}, x \notin A \cup B$, and give $Y=X / R$ the quotient topology wrt the canonical map $q: X \to X/R$ sending $x$ to its class in $X/R$.
Then $q$ is a closed continuous surjective map, $X$ is Hausdorff and $Y$ is not, as we cannot separate the points $A$ and $B$ of $X/R$, or their inverse images would have separated $A$ and $B$ which was impossible.
We cannot strengthen the example further, as the closed continuous image of a $T_4$ (normal and $T_1$) space is again $T_4$, so a fortiori Hausdorff.