You are essentially trying to use the fact that if $f\colon G\to K$ is an isomorphism, and $N\triangleleft G$, then $G/N$ is isomorphic to $K/f(N)$. But you have no warrant for assuming that the image of $\mathbb{Z}$ under a putative isomorphism $f\colon \mathbb{Q}\to\mathbb{R}$ would necessarily be equal to $\mathbb{Z}$. In fact, you cannot make such an assertion, because it is false even in $\mathbb{Q}$: there are isomorphisms from $\mathbb{Q}$ to itself _as an abelian group_ that does not map $\mathbb{Z}$ to $\mathbb{Z}$ (e.g., $q\mapsto \frac{q}{2}$).
So your argument cannot work as written. Now, if you could show that for any infinite cyclic subgroup $C$ of $\mathbb{R}$ you have $\mathbb{R}/C$ not torsion, then you'd be done. But the cardinality argument is by far the simplest here.