Every finite subgroup of $\mathbb{Q}/\mathbb{Z}$ is cyclic

Not in the direction you wish but you can identify $\mathbb{Q}/\mathbb{Z}$ as a subgroup of the complex numbers: $\mathbb{Q}/\mathbb{Z} < \mathbb{R}/\mathbb{Z} \cong S^1 \subset \mathbb C^\times$, and then use that every finite multiplicative subgroup of a field is cyclic.

Following your characterisation of cyclic groups it suffices to show that $\mathbb{Q}/\mathbb{Z}$ contains only $p-1$ elements of order $p$ for any given prime $p$. Those are easy to find and it is straightforward to show that there cannot be any more elements of this order.

Consider the natural homomorphism $\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}$ and lift a set $\overline{q}_1,\ldots ,\overline{q}_r$ of generators of the finite subgroup $\overline{G}$ under consideration to $\mathbb{Q}$. The group $\overline{G}$ then is the image of $G:=\mathbb{Z}q_1 +\ldots \mathbb{Z}q_r$, where $q_i$ is the preimage of $\overline{q}_i$. The $\mathbb{Z}$-module $G$ is free of rank $1$ - one only needs the existence of gcd's in $\mathbb{Z}$ to see this. Thus $G$ and hence $\overline{G}$ are cyclic.