Sorry! The answer doesn't seem obvious at all to me...
If $\langle S \rangle \leqslant \langle T \rangle$ can $S$ be infinite while $T$ finite? I think the answer is yes.
Let $\Bbb{Z}^{\times} \supset T = \{p\}$. Then choose $p^2, p^3, p^5, p^7, \dots, p^{\text{prime}}$ for $S$.
Your counter example seems good to me. Edit (following @bof) : actually it is not that good. If we have $H=\langle p^q\mid q\text{ is prime} \rangle$ then $H=\langle p^2,p^3\rangle$. The reason is the following, clearly $H$ will contain it. Furthermore, $p^2,p^3,p^4,p^5$ are in $H$. And if $n\geq 6$ at least one $p^n$ or $p^{n-2}$ or $p^{n-4}$ will be something like $p^{3k}$ so that any element in $H$ will be contained in $\langle p^2,p^3\rangle$.
Remark that such counter-example also exist for groups.
The classical counter-example is $\mathbb{F}_2=\langle a,b\rangle$ with two generators and its derived subgroup $D(\mathbb{F}_2)$, it can be shown that $D(\mathbb{F}_2)$ is not finitely generated (it can be shown that this is the fundamental group of an infinite grid).