 Solution

Suppose there were only finitely many prime numbers, say $p_1,\dots ,p_k$. Work in the infinite group $(\mathbb{Z},+)$ and set $H_i=p_i\mathbb{Z}$. The left cosets are the $p_i$ residue classes modulo $p_i$, so each $H_i$ is a finite-index subgroup, with $|\mathbb{Z}/H_i|=p_i$.

By part (b)(v), $\mathbb{Z}\setminus {\bigcup }_i{p}_i\mathbb{Z}$ is empty or infinite. But part (a)(ii) identifies it as the nonempty two-element set $\{-1,1\}$. This contradiction proves Euclid's theorem: