Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $\langle r \rangle = H$. If some $s$ in $G$ is not in $H$ then $\langle s \rangle \neq H$. So for all $s$ in $G$, $\langle s \rangle \neq H$ Thus for all $s$ in $G$ there exists an $h\in H$ such that $h \notin \langle s \rangle$. But this is a contradiction with the fact that $a$ generates $G$.
I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).