Why is this quotient space not Hausdorff?

I am trying to show that the following space is not Hausdorff. Consider the topological space $S^1$, and let $r$ be an irrational number. Consider the action of $\mathbb{Z}$ on $S^1$ given by $$ S^1\times\mathbb{Z}\to S^1; (e^{ix}, n)\mapsto e^{i(x+2\pi n r)}. $$ Let $S^1/\mathbb{Z}$ denote the orbit space. I want to show that this space is not Hausdorff.

It was suggested to me that I try showing that any orbit under this action is dense in $S^1$, but I am getting stuck on proving that bit. But here is what I was thinking: we know that the topological group $\mathbb{R}/2\pi \mathbb{Z}$ is homeomorphic to $S^1$, as seen by the map $t\mapsto e^{it}$. Denote the following composition of maps $$ \mathbb{R}\to \mathbb{R}/2\pi \mathbb{Z}\simeq S^1\to S^1/\mathbb{Z} $$ by $\phi$. Then if $[e^{ix}]\in S^1/\mathbb{Z}$, it follows that $$ \phi^{-1}([e^{ix}]) = \{x+2\pi(nr+m)\mid n, m\in\mathbb{Z}\}. $$ If I can show that this subset is dense in $\mathbb{R}$, then it follows that the set $[x]$ as a subset of $S^1$ is dense. This is where I am getting stuck, and it is not even clear to me that this is necessarily true.

Any hints or suggestions are appreciated. Thanks!

It suffices to show that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$, where $x\bmod 1=x-\lfloor x\rfloor$. This is true precisely in case $r$ is irrational. That it’s false for rational $r$ is obvious, so assume that $r$ is irrational. Let $m$ be any positive integer. By the pigeonhole principle there must be distinct $i,j\in\{1,\dots,m+1\}$ and $k\in\{0,\dots,m-1\}$ such that $\frac{k}m\le ir\bmod 1,jr\bmod 1<\frac{k+1}m$. Then $|(j-i)r|\bmod 1<\frac1m$, so every point of $[0,1)$ is within $\frac1m$ of the set $\{n(j-i)r\bmod 1:n\in\Bbb Z\}$, and it follows immediately that $\{nr\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$.