$\mathbb{R}\mathrm{/}\mathbb{Z}$ is Hausdorff

Choose representatives $x,y\in \mathbb{R}$ of $\xi ,\eta$ and put $a=x-\lfloor x\rfloor$, $b=y-\lfloor y\rfloor$. By [L2], $a,b\in [0,1)$; by [L4], $[a]=\xi$ and $[b]=\eta$. Distinctness gives $a\mathrm{\ne }b$, so $d:=\mathrm{\mid }a-b\mathrm{\mid }$ and $e:=1-d$ are both positive.

Put $r=\frac{1}{3}\min \{d,e\}>0$, and let $U=p[(a-r,a+r)]$ and $V=p[(b-r,b+r)]$. Both are open by [L1], and they contain $\xi$ and $\eta$, respectively.

Suppose $U\cap V\mathrm{\ne }\mathrm{\varnothing }$. Then some $u\in (a-r,a+r)$ and $v\in (b-r,b+r)$ have $p(u)=p(v)$, so $k:=u-v\in \mathbb{Z}$ by [L4] and $\mathrm{\mid }k-(a-b)\mathrm{\mid }<2r$. But $a-b\in (-1,1)\setminus \{0\}$, and its distance from every integer is at least $\min \{\mathrm{\mid }a-b\mathrm{\mid },1-\mathrm{\mid }a-b\mathrm{\mid }\}=\min \{d,e\}>2r$: the candidates $0$ and the nearer of $1,-1$ give those two distances, while every other integer is farther away. This is a contradiction. Thus $U$ and $V$ are disjoint open neighbourhoods, and [L3] proves the Hausdorff condition.