The square with opposite edges identified is homeomorphic to the product $(\mathbb{R}/\mathbb{Z}) \times (\mathbb{R}/\mathbb{Z})$

$Q$ is continuous, its components $q \circ \pi_0$ and $q \circ \pi_1$ being composites of continuous maps, and surjective, since every $(a,b) \in T\times T$ is $(q(u),q(v))$ for some $u,v$ by surjectivity of $q$.

For $U, V \subseteq \mathbb{R}$: $Q[U \times V] = q[U] \times q[V]$. Indeed $Q(u,v) = (q(u),q(v))$ gives the inclusion $\subseteq$, and conversely $(q(u), q(v))$ with $u \in U$, $v \in V$ is $Q(u,v)$.

$Q(x,y) = Q(x',y')$ exactly when $x - x' \in \mathbb{Z}$ and $y - y' \in \mathbb{Z}$, by [A1] applied in each coordinate.

$E = Q \restriction S$ is continuous by [L3], and surjective: given $(a,b) \in T \times T$ write $a = q(u)$, $b = q(v)$; then $u - \lfloor u \rfloor$ and $v - \lfloor v \rfloor$ lie in $[0,1)$ by [A2] and $E(u - \lfloor u\rfloor,\ v - \lfloor v \rfloor) = (a,b)$ by [A1].

Fix integers $m,n$ and put $R := [m-1,\ m+1] \times [n-1,\ n+1]$. For $\alpha,\beta \in \{0,1\}$ let $R_{\alpha\beta} := [m-1+\alpha,\ m+\alpha] \times [n-1+\beta,\ n+\beta]$ and define $g_{\alpha\beta} : R_{\alpha\beta} \to S/\!\sim$ by $g_{\alpha\beta}(x,y) := P\big(x - (m-1+\alpha),\ y - (n-1+\beta)\big)$. Each $g_{\alpha\beta}$ is continuous, being $P$ composed with a translation of each coordinate into $[0,1]$, which is continuous by [L2], [L3] and [L5].

For $(s,t),(s',t') \in S$: $E(s,t) = E(s',t')$ exactly when $(s,t) \sim (s',t')$, by step 1.3 and the definition of $\sim$. So the fibres of $E$ are exactly the classes of $\sim$.

$F$ is constant on the fibres of $Q$: if $Q(x,y) = Q(x',y')$ then $x' = x + m$ and $y' = y + n$ for integers $m,n$ by step 1.3, and then $x' - \lfloor x'\rfloor = x - \lfloor x \rfloor$ and $y' - \lfloor y' \rfloor = y - \lfloor y \rfloor$ by [A2].

$Q$ is an open map: by [L1] and [L2] the boxes $U \times V$ with $U,V$ open in $\mathbb{R}$ form a basis of $\mathbb{R}^2$, their images are the boxes $q[U] \times q[V]$ by step 1.2, which are open in $T \times T$ by [A1] and [L1], and the image of a union is the union of the images. With step 1.1 and [L4] this makes $Q$ an open quotient map, which is claim 1.

The four maps of step 1.5 agree on the overlaps of the $R_{\alpha\beta}$, which are contained in the lines $x = m$ and $y = n$. On $x = m$ the two candidate values differ only in that the first coordinate of the argument of $P$ is $1$ in one and $0$ in the other, and $(1,u) \sim (0,u)$; on $y = n$ the same holds in the second coordinate, and at $(m,n)$ all four values are $P$ of the four corners of $S$, which are all $\sim$-equivalent.

By steps 1.5 and 2.4 and the finite closed cover $\{R_{\alpha\beta}\}$ of $R$, [L3] gives a continuous $g : R \to S/\!\sim$ restricting to each $g_{\alpha\beta}$; and $g = F$ on $R$, since for $x \in [m-1,m)$ one has $\lfloor x \rfloor = m-1$, for $x \in [m,m+1)$ one has $\lfloor x \rfloor = m$, and at $x = m+1$ the value $P(\dots,\ 0,\dots)$ agrees with $P(\dots, 1, \dots)$ by $\sim$, the same three cases applying to $y$.

$F$ is continuous: the open sets $(m-1,m+1) \times (n-1,n+1)$, $m,n \in \mathbb{Z}$, cover $\mathbb{R}^2$ by [L1] and [L2], and on each of them $F$ is the restriction of the continuous $g$ of step 3.1, hence continuous by [L3]; the open cover clause of [L3] then gives continuity of $F$.

By step 2.1 and [L4] applied to the quotient map $P$ and the continuous map $E$ of step 1.4, there is exactly one continuous $\bar E : S/\!\sim\ \to T \times T$ with $\bar E \circ P = E$; by step 2.2 and [L4] applied to the quotient map $Q$ of step 2.3 and the continuous $F$ of step 4.1, there is exactly one continuous $\bar F : T \times T \to S/\!\sim$ with $\bar F \circ Q = F$.

$\bar F \circ \bar E = \mathrm{id}$: for $(s,t) \in S$ one has $\bar F(\bar E(P(s,t))) = \bar F(Q(s,t)) = F(s,t) = P(s - \lfloor s\rfloor,\ t - \lfloor t \rfloor)$, which equals $P(s,t)$ in every case, since $\lfloor u \rfloor = 0$ for $u \in [0,1)$ and $1 - \lfloor 1 \rfloor = 0$ with $(0,\cdot) \sim (1,\cdot)$ and $(\cdot,0) \sim (\cdot,1)$; and $P$ is surjective.

$\bar E \circ \bar F = \mathrm{id}$: for $(x,y) \in \mathbb{R}^2$ one has $\bar E(\bar F(Q(x,y))) = \bar E(F(x,y)) = E(x - \lfloor x\rfloor,\ y - \lfloor y\rfloor) = Q(x - \lfloor x \rfloor,\ y - \lfloor y \rfloor) = Q(x,y)$ by [A1] and [A2]; and $Q$ is surjective by step 1.1.

By steps 6.1 and 6.2 the continuous maps $\bar E$ and $\bar F$ are mutually inverse, so $\bar E$ is a homeomorphism, which is claim 2; with step 2.3 both claims are proved.