Is my proof to the question correct?
In $\mathbb{C}$ we have that $i$ is the solution to $x^2 + 1 = 0 $. Thus if a homomorphism exists from $\mathbb{C} \to \mathbb{R}$ there is a solution in $\mathbb{R}$ to
$ ϕ(x^2 + 1) = ϕ(0)$
$ ϕ(x^2)+ ϕ(1) = ϕ(0)$ by closure under addition property of homomorphisms
$ ϕ(x^2) + 1 = 0 $ since $ϕ(1) = 1$ and $ϕ(0) = 0 $ in all ring homomorphisms
$ (ϕ(x))^2 = -1$
which there isn't. Thus no ring homomorphism of this type exists.
In your proof, you use that $\phi(1)=1$ for any ring homomorphism, whether this is true depends on your definition for those (a matter of convention). As you stated this fact, I assume that this is the setting you are working with, in which case your proof is correct.
In the more general case, $\phi(1)=\phi(1\cdot 1) = \phi(1)\cdot \phi(1) = (\phi(1))^2$, which must always be non-negative as $\phi(1)$ is a real number. But then analogously to your proof, $(\phi(x))^2 + \phi(1)=0$ and both are non-negative, so $\phi(x)=0$ for all $x$. Therefore the ring homomorphism $0$ would then be the only one fulfilling your requirements (and it obviously does not send 1 to 1)