An interval of length one need not embed under $p:\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$

The restriction $f=p{\mathrm{\mid }}_{[0,1)}$ is continuous by [L1] and [L3]. It is surjective: for $x\in \mathbb{R}$, [L5] gives $r=x-\lfloor x\rfloor \in [0,1)$ with $p(r)=p(x)$. It is injective: if $r,s\in [0,1)$ and $p(r)=p(s)$, then [L1] gives $r-s\in \mathbb{Z}$ and $\mathrm{\mid }r-s\mathrm{\mid }<1$, so $r=s$. Thus $f$ is a continuous bijection onto $\mathbb{R}\mathrm{/}\mathbb{Z}$.

The set $A=[0,1\mathrm{/}2)$ is relatively open in $[0,1)$, since $A=(-1\mathrm{/}2,1\mathrm{/}2)\cap [0,1)$ by [L4]. Its image satisfies $p^{-1}(p[A])={\bigcup }_{n\in \mathbb{Z}}[n,n+1\mathrm{/}2)$ by [L1]. This union is not open at any integer, in particular at $0$, so [L2] says $p[A]$ is not open in the quotient. Hence the continuous bijection $f$ is not open and is not a homeomorphism by [L6].

The other endpoint convention fails differently: on $[0,1]$ one has $p(0)=p(1)$ by [L1], so the restriction is not injective and cannot be a homeomorphism onto its image. Both intervals have length one, which refutes the proposed replacement of the strict bound in [L7] by length at most one.