= Solution
The <set> $X/\mathcal R$ consists of the <equivalence classes> $[x]$ for $\mathcal R$. The canonical <surjection> $q:X\to X/\mathcal R$, $q(x)=[x]$, defines its <quotient topology>:
$$
O\subseteq X/\mathcal R\text{ is open}\quad\Longleftrightarrow\quad q^{-1}(O)\text{ is open in }X.
$$
The <quotient space> includes this <topology>, not just the collection of <equivalence classes>.
Regard the <circle> as $S^1=\{z\in\mathbb C:|z|=1\}$ and use the <continuous> <surjection>
$$
h:S^1\longrightarrow[0,1],\qquad h(z)=\frac{1+\operatorname{Re}z}{2}.
$$
Define $z\mathcal R w$ precisely when $h(z)=h(w)$. This is an <equivalence relation>, being equality of <function> values. Equivalently $w=z$ or $w=\overline z$, so the construction is the <circle modulo reflection>.
Since $h$ is constant on each <equivalence class>, it induces a <bijection>
$$
\overline h:S^1/\mathcal R\longrightarrow[0,1],\qquad [z]\longmapsto h(z).
$$
It is <continuous> by the <universal property of the quotient topology>, because $\overline h\circ q=h$ is <continuous>. The <circle> is <compact> by the <Heine-Borel theorem>, and its <quotient space> is a <continuous image of a compact space>, hence <compact>. The <closed interval> is <Hausdorff>. Part (a)(ii) therefore proves
$$
\boxed{S^1/\mathcal R\cong[0,1].}
$$
Geometrically, the <equivalence relation> folds the upper and lower semicircles together, retaining the two real endpoints separately.
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