I stumbled upon this link while googling for an answer to the same question.
Under point number 3, the sole answer mentions an example where it is said $\mathbb{R}/\mathbb{Q}$ is not Hausdorff, because the topology generated is trivial, assuming $\mathbb{R}$ has the standard topology. I agree with the conclusion but I think the reason given is wrong. I just wanted to cross-check my argument.
If I take $ (\mathbb{R}/\mathbb{Q}) \backslash [e] $ where e is the equivalence class of the Euler number $e$,
then the inverse of the projection map $\pi: \mathbb {R}\rightarrow \mathbb{R}/\mathbb{Q}$ is $\mathbb{R}\backslash \{e\}$ which is an open set in $\mathbb{R}$.
And so the topology consists of open sets other than $X$ and $\phi$, which means the topology is not trivial.
Is there something wrong with the above argument?
NOTE: I agree with the conclusion though. Let $U \neq \phi$ be an open set in $\mathbb{R}/\mathbb{Q}$ . Let $q\in \mathbb{Q}$ then if $[q] \notin U$, $\pi^{-1}(U)$ will consist of only irrationals which cannot be open, and so the space is not hausdorff, since any two non-empty open sets have non-empty intersection containing atleast $[q]$.
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 continuous surjection that is constant on $D$, and suppose that $f$ maps every element of $D$ to $q$.
Let $U \subseteq X$ be a non-empty open set, 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, unless it contains at most one point.