$\mathbb{R}\mathrm{/}\mathbb{Z}$: the quotient map is open, and the quotient is homeomorphic to $[0,1]$ with its endpoints identified

For $c\in \mathbb{R}$ the translation ${\tau }_c(t):=t+c$ carries an open $U$ to an open set: if $y\in U+c$ then $y-c\in U$, so [L1] gives $r_0>0$ with $(y-c-r_0,y-c+r_0)\subseteq U$, whence $(y-r_0,y+r_0)\subseteq U+c$.

For $U\subseteq \mathbb{R}$: $q^{-1}[q[U]]={\bigcup }_{m\in \mathbb{Z}}(U+m)$, since $q(x)=q(u)$ for some $u\in U$ exactly when $x-u\in \mathbb{Z}$, that is $x\in U+m$ for some integer $m$.

$j$ is continuous, being a restriction of the continuous $q$; and $j$ is surjective, since for $x\in \mathbb{R}$ the number $x-\lfloor x\rfloor$ lies in $[0,1)$ by [A2] and satisfies $q(x-\lfloor x\rfloor )=q(x)$.

For $s,t\in [0,1]$: $j(s)=j(t)$ exactly when $s-t\in \mathbb{Z}$, and since $\mathrm{\mid }s-t\mathrm{\mid }\le 1$ that happens exactly when $s=t$ or $\{s,t\}=\{0,1\}$. So the fibres of $j$ are exactly the classes of $\approx$.

$r$ is constant on the fibres of $q$: if $q(x)=q(y)$ then $y=x+m$ for an integer $m$, and $\lfloor y\rfloor =\lfloor x\rfloor +m$ by the uniqueness in [A2], so $y-\lfloor y\rfloor =x-\lfloor x\rfloor$ and $r(y)=r(x)$.

For each integer $m$ define $g_m:[m-1,m+1]\to K$ by $g_m(t):=p(t-m+1)$ for $t\in [m-1,m]$ and $g_m(t):=p(t-m)$ for $t\in [m,m+1]$. The two clauses agree at $t=m$, giving $p(1)$ and $p(0)$, which are equal because $\{0,1\}$ is one class of $\approx$.

By step 1.1 and step 1.2 the saturation of an open $U$ is a union of open sets, hence open; so $q[U]$ is open in $T$ by [A1], and $q$ is an open map, hence an open quotient map by [L5]. This is claim 1.

Each clause of step 1.6 is continuous: $t\mapsto t-m+1$ and $t\mapsto t-m$ are continuous by step 1.1 read through [L1], they map the stated closed interval into $[0,1]$, and $p$ is continuous; so [L2] and [L3] apply. By the finite closed cover $\{[m-1,m],[m,m+1]\}$ of $[m-1,m+1]$ and [L3], $g_m$ is continuous.

$g_m$ agrees with $r$ on $[m-1,m+1]$: for $t\in [m-1,m)$ one has $\lfloor t\rfloor =m-1$ and $r(t)=p(t-m+1)=g_m(t)$; for $t\in [m,m+1)$ one has $\lfloor t\rfloor =m$ and $r(t)=p(t-m)=g_m(t)$; and at $t=m+1$ one has $r(t)=p(0)=p(1)=g_m(t)$.

$r$ is continuous: the open intervals $(m-1,m+1)$, $m\in \mathbb{Z}$, cover $\mathbb{R}$, and on each of them $r$ agrees with a restriction of the continuous $g_m$ by steps 2.2 and 2.3, hence is continuous there by [L2]; [L3] then gives continuity of $r$.

By step 1.4 and [L4] applied to the quotient map $p$ and the continuous $j$, there is exactly one continuous $\bar{j}:K\to T$ with $\bar{j}\circ p=j$; by step 1.5 and [L4] applied to the quotient map $q$ and the continuous $r$ of step 3.1, there is exactly one continuous $\bar{r}:T\to K$ with $\bar{r}\circ q=r$.

$\bar{r}\circ \bar{j}={\mathrm{i}\mathrm{d}}_K$: for $t\in [0,1]$ one has $\bar{r}(\bar{j}(p(t)))=\bar{r}(j(t))=\bar{r}(q(t))=r(t)=p(t-\lfloor t\rfloor )$, which is $p(t)$ for $t\in [0,1)$ and $p(0)=p(1)$ for $t=1$; so $\bar{r}\circ \bar{j}\circ p=p$, and $p$ is surjective.

$\bar{j}\circ \bar{r}={\mathrm{i}\mathrm{d}}_T$: for $x\in \mathbb{R}$ one has $\bar{j}(\bar{r}(q(x)))=\bar{j}(r(x))=\bar{j}(p(x-\lfloor x\rfloor ))=j(x-\lfloor x\rfloor )=q(x-\lfloor x\rfloor )=q(x)$; so $\bar{j}\circ \bar{r}\circ q=q$, and $q$ is surjective.

By steps 5.1 and 5.2 the maps $\bar{j}$ and $\bar{r}$ are mutually inverse, and both are continuous by step 4.1; so $\bar{j}$ is a homeomorphism $K\to T$, which is claim 2. With step 2.1 both claims are proved.