# The diagonal in $T_1$ space. [duplicate]

Hi¡ I have some troubles with the next problem.

Let $(X,\tau)$ be a topological space. Prove that $X$ is $T_1$ if and only if there exists a family $U$ of open sets such that $\bigcap U=\Delta$ where $\Delta=\left\{(x,x): x\in X\right\}\subseteq X\times X$ (the diagonal)

My attempt.

$\Rightarrow)$ We know that $X\times X$ is $T_1$ because $X$ is $T_1$. Moreover, we have the next theorem for $T_1$ spaces

Theorem Let $X$ be a topological space. The next conditions are equivalent.

1) $X$ is $T_1$

2) For all $B\subseteq X$, $B=\bigcap \left\{ U: B\subseteq U, U\in\tau\right\}$

3) For all $x\in X$, $\left\{x\right\}=\bigcap\left\{U : U\in\tau, x\in U\right\}$

Then, the implication follows from 3).

But, what can I do for $\Leftarrow)$? I have tried to prove that $\left\{x\right\}$ is closed, but I have failed. My best attempt was consider two distinct points $x$ and $y$. Clearly, $(x,y)\notin\Delta$, then, $(x,y)\notin\bigcap U$, so, there exist some basic open set such that $(x,y)\notin A\times B$. Then, my idea was use the projection $\Pi_X$, but, from here, I'm so confused.