How to show this?

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> Let $(X,\cal T)$ be a topological space. For $x\in X$, let $\cal U$ be the family of all open neighborhoods of $x$. Show that: if $\bigcap\limits\_{U\in\cal U}\overline U=\{x\}$, for all $x\in X$, then $X$ is Hausdorff.
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Here $\overline U$ is closure of $U$.