Error in proof: $\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$??

It's not too tough to spot! $i\notin\mathbb R$.

What your proof does in fact show is that by Chinese Remainder, $\mathbb C[X]/(X^2+1)\cong \mathbb C\times\mathbb C$.

The kernel of your homomorphism is $(X^2+1)$, not $(X-i)$, since $X-i\notin\mathbb R[X]$.

What does $\mathbb{R}[X]/(X+\operatorname{i})$ mean? The polynomial $X+\operatorname{i}$ does not belong to $\mathbb{R}[X]$, and so $\mathbb{R}[X]/(X+\operatorname{i})$ does not make sense. This is because $\operatorname{i}$ does not belong to $\mathbb{R}$.

$ \Bbb R[x]/(x+i)$ is not defined as $x+i$ is not even in $ \Bbb R[x]$.