Suppose H is an arbitrary subgroup of G and $H\neq \{e\}$. Let $a^{k_{0}}\in H$. Let $b=a^{t}$ be an arbitrary member of H such that $k_{0}\le t$. For some $q,r$ s.t. $t=k_{0}q+r$, then $a^{t}=a^{k_{0}q+r}=a^{k_{0}q}a^{r}$ for $0\le r\lt k_{0}$. Since $k_{0}\le t$, $r=0$. Thus, $a^{t}=a^{k_{0}q}$ and therefore H is cyclic. Since H was arbitrary, any subgroup of a cyclic group is also cyclic.
Is this proof correct? Any possible improvements welcome
Hint: It would be helpful if you start with the integer $k_0$, which is the least one such that $a^{k_0}\in H$. This would a generator of $H$.