$\mathbb{R}/\mathbb{Z}$: 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 $|s-t| \le 1$ 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 $\bar r \circ q = r$.

$\bar r \circ \bar j = \mathrm{id}_K$: 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.

$\bar j \circ \bar r = \mathrm{id}_T$: 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.