Let $\mathbb{I} = \mathbb{R}\backslash\mathbb{Q}$ be the set of irrational numbers. Define the equivalence relation $\sim$ on $\mathbb{I}$ such, that $\forall x,y\in\mathbb{I}:\big((x\sim y) \Leftrightarrow (x - y \in \mathbb{Q})\big)$. Is $\mathbb{I}/{\sim}$ either compact, connected, Hausdorff?
As I understand it is not Hausdorff, because any neighborhood of point $x$ in $\mathbb{I}$ contains some points which are located on enough small rational distance from $x$. Therefore any neighborhood of $[x]_{\sim}$ is intersecting with the neighborhood of some other points.
The space $\mathbb{I}$ itself with induced topology is not connected or compact, however it does not entail that $\mathbb{I}/{\sim}$ is not compact or connected.
I’ve tried to work with quotient metric, but it is trivial here. So i’m stuck a little bit.
On connectedness and compactness:
Consider any two open sets ($\mathbb{I}$-wise) $X_{1},X_{2}\in \mathbb{I}$. Let \begin{equation} q: \mathbb{I}\to\mathbb{I}/{\sim} \end{equation} be a quotient map. Note, that there always exist such $x_{1}\in X_{1}$ and $x_{2}\in X_{2}$, that $x_{1} - x_{2}\in\mathbb{Q}\Rightarrow x_{1}\sim x_{2}$. Then $q(X_{1})\cap q(X_{2})\ne 0$. That means that $\mathbb{I}$ could not be represented as two open disjoint sets and, therefore, $\mathbb{I}/{\sim}$ is (topologically) connected (by definition).
Moreover note that the same argument tells us that the representatives of all of the equivalency classes $[x]_{\sim}$ are contained in any arbitrary small interval of $\mathbb{R}$. That entails that the quotient topology on $\mathbb{I}/{\sim} $ is indiscrete. And therefore all of the open coverings of $\mathbb{I}/{\sim} $ are finite, which of course mean that $\mathbb{I}/{\sim}$ is compact.