= Solution
The difference map $(z,w)\mapsto z-w$ is <continuous>, and $X$ is its <preimage> of the <open set> $\mathbb C\setminus\{0\}$. Thus $X$ is open in the product.
The class of $(z,w)$ is $C_z=\{z\}\times(\mathbb C\setminus\{z\})$. A punctured plane is <path-connected>: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is <connected>.
The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is <continuous> as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-<topology> <basis> gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq\pi(U)$. Hence $\pi(U)$ is open.
The map is constant exactly on the equivalence classes, so it induces a bijection $\overline\pi:X/R\to\mathbb C$. It is <continuous> by the defining property of the <quotient topology>. It is also open: for quotient-open $V$, $\overline\pi(V)=\pi(q^{-1}(V))$ is open. Thus
$$
\boxed{X/R\cong\mathbb C\text{ homeomorphically}}.
$$
The plane is <connected>, and the classes are <connected>, so part (b) now yields \b[$X$ is <connected>].
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