Every hausdorff space has a non-hausdorff quotient. I know some Hausdorff spaces can have non-hausdorff quotient spaces. For example real line with double origin.
I am wondering whether this is the case for all Hausdorff spaces. I tried some simpler examples like for a given Hausdorff space I tried to write a quotient map to a finite set with indiscrete topology but couldn't manage to find such map for an arbitrary Hausdorff space either.
At this point, I have no clue about how to approach showing whether a Hausdorff space without a non-Hausdorff space exits or not.
So in short, the question is written in the title. Does every Hausdorff space have a non-Hausdorff quotient space?
 A: Let $X$ be any non-empty set with the discrete topology. Let $\sim$ be any equivalent relation on $X$. Let $X/\sim$ be given the quotient topology with the quotient map $\pi : X\to X/\sim$. Then $Y\subset X/\sim$ is open iff $\pi^{-1}(Y)$ is open: that is, the quotient topology on $X/\sim$ is also the discrete topology and thus is Hausdorff.
A: First of all note that a quotient of a discrete space is discrete (thus Hausdorff).
But the answer is "yes", if $X$ is any non-discrete space (Hausdorff or not, irrelevant). Since $X$ is not discrete then there is a non-isolated point $x_0\in X$. Define equivalence relationship on $X$ by $x\sim y$ if $x=y=x_0$ or $x\neq x_0$ and $y\neq x_0$. In other words we collapse $X\backslash\{x_0\}$ to a point. Note that the quotient $X/\sim$ is not Hausdorff. Since the quotient is finite (it has only two points) then being Hausdorff is equivalent to being discrete. But $\{[x_0]\}$ is not open in $X/\sim$ because that would mean that $\{x_0\}$ is open in $X$ (as the preimage under quotient map), but $x_0$ is not isolated.
