@MattN The kernel of the homomorphism $x \mapsto e^{ix}$ is $2\pi \mathbb{Z}$. Now the kernel of the homomorphism $x \mapsto \varphi(e^{i x})$ is a closed subgroup of $\mathbb{R}$ and it contains $2\pi \mathbb{Z}$. Moreover $x \mapsto \varphi(e^{ix})$ is of the form $x \mapsto e^{\lambda i x}$ for some $\lambda \in \mathbb{R}$ because you know what the homomorphisms $\mathbb{R} \to S^1$ are. What possibilities for $\lambda$ do you have?
@MattN There's nothing complicated about this! Your proof goes wrong because you claim something that you don't justify in the middle of it (just before "We claim"). I'm providing the reason for it.
