Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic

A nicer proof, perhaps, is to note that if $\Bbb Q^\times$ were cyclic, being infinite, must isomorphic to $\Bbb Z$. But $\Bbb Z$ has no element of order $2$, whereas $(-1)^2=1$ in $\Bbb Q^\times$. Note that this proves then that $\Bbb R^\times$ cannot be cyclic either.

ADD To be more precise, $$\Bbb Q^\times \simeq \Bbb Z/2\Bbb Z \oplus \bigoplus_{i\geqslant 1}\Bbb Z$$ by using the prime factorization.

Hint: If ${\mathbb R}$ is cyclic then it's countable. Your argument for ${\mathbb Q}$ looks good. Also a subgroup of a cyclic group is cyclic so you could go that route for ${\mathbb R}$ too, based on your proof for ${\mathbb Q}$.

Simple proof:

If $|x|>1$, then $|x^n| > 1$ for all $n \in \mathbb{N}$.

If $|x|<1$, then $|x^n|< 1$ for all $n \in \mathbb{N}$.

So take a purported generator $g$. Well, $|g| \neq 1$, so either $|g|<1$, or $|g|>1$. In the former, we won't generate any numbers with magnitude larger than $1$. In the latter, we will not generate any numbers with magnitude less than $1$. Thus, neither group can be cyclic.