If $X$ is Hausdorff and $\sim$ is an equivalence relation in $X$, then $X/\sim$ endowed with the quotient topology is also Hausdorff

To make this problem I am using the following post:

$X/\sim$ is Hausdorff if and only if $\sim$ is closed in $X \times X$

So, I would say that this is false since the equivalence relation must be closed, how can I find a counterexample of this? Could anyone help me please? Where where $X$ is Haudorff but $X/\sim$ is not. Thank you very much.

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    $\begingroup$ Let $X = \mathbb{R}$, and for $\sim$ take something having to do with $\mathbb{Q}$. That has a good chance to work. $\endgroup$ – Daniel Fischer Oct 20 '17 at 20:19
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    $\begingroup$ I think that if you take $\Bbb R\times\Bbb R$ for $X$, and $(x,y)\sim(x',y')$ when there is positive real $\lambda$ with $x'=\lambda x$ and $y'=\lambda y$, then you get something that’s the circle plus one point $\Bbb O$ for the origin. Clearly not Hausdorff. $\endgroup$ – Lubin Oct 20 '17 at 20:23

Consider the line with two origins, where we take two disjoint copies of $\mathbb{R}$ and identify all the obvious pairs of points except the two origins.

That is, $X = \mathbb{R} \times \{ 0, 1 \}$, where $(x,0)$ is equivalent to $(x,1)$ precisely for $x \ne 0$. Then all neighbourhoods of $(0,0)$ intersect all neighbourhoods of $(0,1)$, so $X$ is non-Hausdorff. See the Wikipedia link for further discussion.

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