The quotient map is open, and every interval shorter than one embeds in $\mathbb{R}\mathrm{/}\mathbb{Z}$

A real $x$ lies in $p^{-1}(p[U])$ exactly when $p(x)=p(u)$ for some $u\in U$, which by [L1] is equivalent to $x-u=n$ for some $n\in \mathbb{Z}$; hence $p^{-1}(p[U])={\bigcup }_{n\in \mathbb{Z}}(U+n)$. Each translate $U+n$ is open: translating by $n$ and by $-n$ gives mutually inverse continuous maps by [L4]. The union is open, so the quotient-topology criterion in [L1] makes $p[U]$ open. Thus $p$ is open in the sense of [L3], including when $U=\mathrm{\varnothing }$.

Suppose $x,y\in J$ and $p(x)=p(y)$. Then $k:=x-y\in \mathbb{Z}$ by [L1], while $\mathrm{\mid }k\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\ell }<1$. If $0<k<1$, both $0$ and $k$ satisfy the integer-part inequalities for the real $k$, contrary to uniqueness in [L2]; applying the same argument to $-k$ excludes $-1<k<0$. Hence $k=0$ and $x=y$, so $p{\mathrm{\mid }}_J$ is injective. This also covers a singleton interval; for an empty interval injectivity is vacuous.

The restriction $p{\mathrm{\mid }}_J$ is continuous and is a bijection onto $p[J]$ by step 1.2. To prove its inverse continuous, let $O$ be relatively open in $J$ and $x\in O$. Choose $\delta >0$ with $J\cap (x-\delta ,x+\delta )\subseteq O$, and put $r=\frac{1}{2}\min \{\delta ,1-\mathrm{\ell }\}>0$. Step 1.1 makes $W:=p[(x-r,x+r)]$ open. If $z\in J$ and $p(z)\in W$, choose $y\in (x-r,x+r)$ with $p(z)=p(y)$; then $z-y\in \mathbb{Z}$ by [L1] and $\mathrm{\mid }z-y\mathrm{\mid }\le \mathrm{\mid }z-x\mathrm{\mid }+\mathrm{\mid }x-y\mathrm{\mid }<\mathrm{\ell }+r<1$, so [L2] gives $z=y$. Thus $z\in J\cap (x-\delta ,x+\delta )\subseteq O$, and $W\cap p[J]\subseteq p[O]$. Every point of $p[O]$ therefore has a relative open neighbourhood contained in $p[O]$, so $p[O]$ is open in $p[J]$. The empty case has the unique empty inverse. Hence $p{\mathrm{\mid }}_J$ is a homeomorphism onto its image, and therefore an embedding by [L3].