Proving $(\Bbb R,+)$ has no proper subgroup isomorphic to itself

There is if you accept AC.

$\Bbb R$ is a vector space over $\Bbb Q$, so has a vector space basis (Hamel basis) as a vector space over $\Bbb Q$. Removing a generator, the remaining generators generate a proper $\Bbb Q$-subspace isomorphic to $\Bbb R$ as a vector space, and so as an Abelian group.

Consider $\mathbb{R}$ as a vector space over the rationals and some basis $B$. Then $B$ is Infinite. Let $b\in B$. Then there is a bijection between $B$ and $B':=B\setminus\{b\}$ implying that the vector spaces generated by $B$ and $B'$ are isomorphic. Thus they are also isomorphic as abelian groups. Finally note that the space generated by $B'$ is strictly contained in the space generated by $B$ which is $\mathbb{R}$.