Proper Subgroup of $O_2(\mathbb{R})$ Isomorphic to $O_2(\mathbb{R})$

$\newcommand{\bZ}{\mathbf Z}\newcommand{\bR}{\mathbf R}$Note that $O(2)\cong (\bR/\bZ)\rtimes (\bZ/2\bZ)$. Thus, it suffices to find a proper subgroup of the torus $\bR/\bZ$ (isomorphic to the torus).

To do this, just fix a basis $B$ of the reals containing $1$ (over the rationals). Then for every $B'\subseteq B$ containing $1$, having the cardinality of the continuum, you have $\operatorname{span}(B')/\bZ\cong \bR/\bZ$: just fix a bijection $B'\to B$ (fixing $1$), extend it to a linear isomorphism $\operatorname{span}(B')\to \mathbf R$ and note that it induces the desired isomorphism.