Suppose $X$ is compact Hausdorff space and $f : X \to Y$ be a quotient map.
Then it is well known that $Y$ is Hausdorff iff $\Delta =\{(x,y) \mid f(x)=f(y) \}$ is closed in $X \times X$. For example a proof of this can be found here. I am looking for a counterexample if $X$ is not compact i.e.
Give an example (preferably non artificial) of a non compact Hausdorff space $X$ such that $\Delta$ is closed in $X \times X$ but $Y$ is not Hausdorff.
As suggested by t.b. in the comments linked to by Justin, you can take any Hausdorff space $X$ which is not regular and thus contains a point $x$ that cannot be separated from some closed set $A$ in $X$. If $q: X \to X/A$ is the quotient map, then the equivalence relation induced by $q$ is the set $\Delta_X \cup (A\times A)$ in $X\times X$, which is closed as $X$ is Hausdorff and $A$ is closed. However, $X/A$ is not Hausdorff, since for $\{x\}$ and $A$ to have disjoint neighborhoods, there would have to be disjoint neighborhoods around $x$ and $A$ in $X$.
Note that for any map $q: X \to Y$, the preimage of $\Delta_Y \subseteq Y\times Y$ under $q\times q$ is the relation $\{(x,x') \mid q(x)=q(x')\}$. If this set is closed and $q\times q$ is a quotient map, then $\Delta_Y$ is closed in $Y \times Y$. Since in the example above this is not the case, we also have a situation where the square $q\times q$ of the quotient map $q$ is not a quotient map.
For a concrete example, consider the space $X$ whose underlying set is $\Bbb R$ and whose topology is the smallest one which contains the Euclidean topology and has $K = \{1, 1/2, 1/3, \dots\}$ as a closed set. The open sets in $X$ are of the form $U \cup V\setminus K$, where $U$ and $V$ are open in the standard topology. Since for any open set $U$ around $0$ and any open set $V$ around $K$, $U\setminus K$ intersects $V$, the space is not regular (Also see my answer for The topology on $X/{\sim}\times X/{\sim}$ is not induced by $\pi\times\pi$.)
