= Solution
A <quotient topological space> of a <topological space> $X$ is obtained by an <equivalence relation> $\sim$: its points are the classes $X/{\sim}$, and the canonical surjection is $q(x)=[x]$. The <quotient topology> declares a subset $U$ of the quotient open exactly when $q^{-1}(U)$ is open in $X$. More generally a surjection onto a set specifies the same topology by this condition.
For a Hausdorff source with a non-Hausdorff quotient, use $X=\mathbb R$ and $x\sim y$ when $x-y\in\mathbb Q$. Any nonempty open inverse image is a nonempty open subset of $\mathbb R$ invariant under rational translations. It contains an interval $I$, and for every real $t$ there is a rational $r$ with $t-r\in I$; hence it contains every $t$. Thus $\mathbb R/\mathbb Q$ has the <indiscrete topology>.
It has distinct points, for instance the classes of $0$ and $\sqrt2$, but neither has a proper nonempty open neighbourhood. \b[The quotient is not Hausdorff], even though $\mathbb R$ is a <Hausdorff space>.
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