= Solution
Define
$$
f:[0,1]^2\longrightarrow S^2,
$$
$$
f(x,y)=\bigl(\sin(\pi y)\cos(2\pi x),
\sin(\pi y)\sin(2\pi x),\cos(\pi y)).
$$
This is continuous and takes values on the <unit sphere>. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the <universal property of the quotient topology> gives a continuous map
$$
\overline f:X/R\longrightarrow S^2.
$$
For $0<y<1$, the third coordinate $\cos(\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline f$ is injective. The spherical-coordinate formula also shows that it is surjective.
The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of $\mathbb R^3$. The <compact-to-Hausdorff continuous bijection theorem> now makes $\overline f$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the <suspension of a topological space> $S^1$, which is $S^2$. This proves the <square quotient model of the two-sphere>.
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