Let $X=\mathbb{R}\cup \{\infty\}$ and for every $p\in \mathbb{R}-\{0\}$, $p$ has the usual neighborhoods in $\mathbb{R}-\{0\}$. Basic neighborhoods of $0$ is the sets of the form $\{[0, \frac{1}{n}):n\in\mathbb{N}\}$ while basic neighborhoods of $\infty$ are the sets of the form $\{(-\frac{1}{n}, 0)\cup\{\infty\}: n\in\mathbb{N}\}$. I think $X$, endowed with this topology, is a Hausdorff space.
Can We say that for every closed neighborhood $N$ of diagonal $X$, $\Delta_X$ , in $X\times X$, $(0, \infty)\in N$?
It's clear that $\overline{U} \cap \overline{V} \neq \emptyset$ whenever $U$ is an open neighbourhood of $0$ and $V$ one for $\infty$. This implies the property you ask about. The above first fact is already a direct proof that $X$ is not Urysohn.