Let $a \in G$ such that $\operatorname{ord}(a)$ is infinite. Then $(a), (a^2), (a^3), \ldots$ are all distinct subgroups of $G$ and hence, $G$ has infinitely many proper subgroups.
Let $G$ be infinite. Then $G$ has infinitely many proper subgroups.
What's the difference between these theorems? Just the possibility that $\operatorname{ord}(a) = n$ for some $n \in \mathbb N$ in the second one?
Yes, that's exactly the point. There are infinite groups, all of whose members have finite order. $\mathbb Q/\mathbb Z$ is an example of such a group.