Example of Hausdorff Space Whose Quotient Space is Not Hausdorff

Solution 1:

The example is fine, but your argument could be simplified a bit: consequently, any open set in the quotient space containing $\{0\}$ necessarily contains $A$, so the quotient not only isn't Hausdorff, it's not $T_1$ (although it is $T_0$, since $\{A\}$ is open).

Solution 2:

Even simpler: $X=\Bbb R$ (usual top.). Let the two classes (for the equivalence relation to induce the quotient) be $\Bbb Q$ and $\Bbb P$ (the irrationals).

Then the only non-empty open set of the quotient space $Y=\{q,p\}$ is $Y$ as all non-empty open subsets of $X$ contain both rationals and irrationals. So $Y$ has the trivial (indiscrete) topology and is not Hausdorff.