This question is similar to Consider $\Bbb R / \Bbb Q = \{ x+\Bbb Q : x \in \Bbb R \}$. Show that the quotient topology is the trivial one. but not the same.
Consider the eq. relation on $\mathbb{R}$ as $x\simeq y \iff x=y$ OR $x,y$ are rational with $x=y+n$ where $n\in\mathbb{Z}$. Let $(X,\mathcal{T}_{\simeq})$ be the quotient space with quotient topology induced by the relation $\simeq$. Is it true that $\mathcal{T}_\simeq$ must also be trivial? Or is it some other topology?
I believe that it is: if $\pi:\mathbb{R}\to X$ is the quotient map and $U\neq\emptyset\subseteq X$ is an open set, then for $\pi^{-1}(U)$ must contain at least one element of all of the rational classes, but then to be open in $\mathbb{R}$ would also have to contain all of the irrational numbers since $\mathbb{Q}$ is dense in $\mathbb{R}$. Any help would be appreciated.
The answer is negative, $\mathscr{T}_\simeq$ is not trivial. I'll just show an example of a non-trivial open set and explain how it came to mind (and some other facts about your space) later. Sin the real line carries the Euclidean topology, we'll refer to it as $E^1$. $X$ will be refered to as $E^1/\simeq$.
The non-trivial open set.
Let $a,b\in(0,1)$ with $a<b$. Let $$G=\bigcup_{n\in\mathbb{Z}}\big\{[x]\in E^1/\simeq\big|\; a+n<x<b+n\big\}=\bigcup_{n\in\mathbb{Z}}\pi\big((a+n,b+n)\big).$$ Notice that $[(a+b)/2]\in G$ and that $[0]\notin G$ so $G\notin\{\emptyset, E^1/\simeq\}$. We have that $$\begin{align*}G&=\bigcup_{n\in\mathbb{Z}}\Big(\pi\big((a+n,b+n)\cap\mathbb{Q}\big)\cup \pi\big((a+n,b+n)\cap\mathscr{C}\mathbb{Q}\big)\Big)\\ &=\bigcup_{n\in\mathbb{Z}}\pi\big((a+n,b+n)\cap\mathbb{Q}\big)\cup \bigcup_{n\in\mathbb{Z}}\pi\big((a+n,b+n)\cap\mathscr{C}\mathbb{Q}\big)\\ &=\pi\big((a,b)\cap\mathbb{Q}\big)\cup \bigcup_{n\in\mathbb{Z}}\bigcup_{\substack{x\in(a+n,\,b+n)\\ x\in \mathscr{C}\mathbb{Q}}}\big\{\{x\}\big\}.\end{align*}$$ Where $\forall x\in\mathscr{C}\mathbb{Q},\; \pi(x)=\{x\}$ was used in the last line. By taking pre-image, $$\begin{align*}\pi^{-1}(G)&=\left(\bigcup_{n\in\mathbb{Z}}(a+n,b+n)\cap\mathbb{Q}\right)\cup\left(\bigcup_{n\in\mathbb{Z}}\bigcup_{\substack{x\in(a+n,\,b+n)\\ x\in \mathscr{C}\mathbb{Q}}}\{x\}\right)\\ &=\left(\bigcup_{n\in\mathbb{Z}}(a+n,b+n)\cap\mathbb{Q}\right)\cup\left(\bigcup_{n\in\mathbb{Z}}(a+n,b+n)\cap\mathscr{C}\mathbb{Q}\right)\\ &= \bigcup_{n\in\mathbb{Z}}(a+n,b+n).\end{align*}$$ Where $\forall x\in\mathscr{C}\mathbb{Q},\; \pi^{-1}\left(\big\{\{x\}\big\}\right)=\{x\}$ was used in the first line. $$\therefore G\in\mathscr{T}_\simeq\setminus\{\emptyset, E^1/\simeq\}.$$
The intuition
Notice that the rational numbers are mapped to the circle (in the abstract sense that $x=x+n=x+2n=\dots$). The only problem with the topology is that $\mathscr{C}\mathbb{Q}$ isn't mapped to the circle; it remains a line. How could one imagine this geometrically? The first thing that comes to mind is a circle (of ratiinal numbers) with a straight line (of irrationals) tangent to it. This intuition fails: consider any sequence $(a_n)_{n\in\mathbb{N}}$ of rational numbers in $(0,1)$ that converge to an irrational number $q_0$ in $(0,1)$. Since $\pi$ is continuous, $\pi(a_n)\to \pi(q_0)=\{q_0\}$. Notice now that, if we let $m\in \mathbb{Z}$, then $a_n+m\to q_0+m$ and $\pi(a_n)=\pi(a_n+m)$, by continuity, again, $\pi(a_n+m)\to\pi(q_0+m)=\{q_0+m\}$. Then $(b_n)_{n\in\mathbb{N}}\colon=\big(\pi(a_n)\big)_{n\in\mathbb{N}}$ is such that $b_n\to \{q_0\}$ and $b_n\to \{q_0+m\}$ (notice btw that since $\{q_0\}\neq \{q_0+m\}$, we've just proven that $E^1/\simeq$ is not Hausdorff) so $\{q_0\}$ and $\{q_0+m\}$ must be close for any irrational $q_0$ and any integer $m$. Our intuition must be re-shaped: Imagine the rational circle with the irrational tangent line again, grab each one of the ends of the tangent line and start forming a sort of spring on both sides of the circle (of same radius). One could imagine the space space $E^1/\simeq$ as this infinite spring with a circle in the middle and say that two elements are close if their angular position is close.
To define $G$, I just took every element of $E^1/\simeq$ that had an angle such that it's projection to the circle in the center was in $(a,b)$.