Countable number of subgroups $\implies $ countable group
Let $S$ be the set of all subgroups of $G$.
Consider the map $\phi: G \to S$ given $\phi(g) = \langle g \rangle$.
If $\phi$ were injective, then we'd be done.
Unfortunately, $\phi$ is not injective, but fortunately we can control how not injective it is.
Indeed, every $H \in S$ has a finite number of pre-images since every cyclic group has finite number of generators (including infinite cyclic groups). Therefore $$ G = \bigcup_{H \in S} \phi^{-1}(H) $$ is a countable union of finite sets and so is countable.