Showing $\mathbb{R}/\mathbb{Q}$ is not Hausdorff

Let's get really broad, and also hope I'm not making any major errors. Forgive (or fix) the awful variable names.

Let $D$ be a dense set in a space $S$.

Let $f\colon S \to X$ be any quotient map that is constant on $D$, and suppose that $f$ maps every element of $D$ to $q$.

Let $y \in X\setminus\{q\}$.

Let $U$ be an open set about $y$, so $f^{-1}[U]$ is a non-empty open set.

Then $f^{-1}[U]$ contains an element, $b$, of $D$, so $q\in U$.

So we see that every non-empty open set in $X$ contains $q$, so $X$ cannot even be $T_1$, let alone Hausdorff.

The topology is trivial: every open set in $\mathbb{R}$ contains a complete set of coset representatives for $\mathbb{R}/\mathbb{Q}$. So if $U\subseteq\mathbb{R}/\mathbb{Q}$ is open and nonempty, $\pi^{-1}(U)$ is open in $\mathbb{R}$ and $\pi(\pi^{-1}(U))$ is all of $\mathbb{R}/\mathbb{Q}$.