Proving $\Bbb C$ is algebraically closed by analytic approach. I want to prove that $\Bbb C$ is algebraically closed in the following way. Let us first take a non-constant polynomial $F$ with coefficients from $\Bbb R.$  Then $F \in \Bbb R[X] \subset \Bbb C[X].$ Now consider the polynomial function $$F : \Bbb C \longrightarrow \Bbb C$$ defined by $z \mapsto F(z).$ If I can prove that $F$ is both open as well as closed in $\Bbb C$ then we are through. Because then $F(\Bbb C) = \Bbb C$ (since $\Bbb C$ is both open and closed in $\Bbb C$ and $\Bbb C$ is connected so either $F(\Bbb C) = \varnothing\ \text {or}\ \Bbb C.$ But since $F$ is a polynomial $F(\Bbb C) \neq \varnothing$). But then $\exists$ $z \in \Bbb C$ such that $F(z)=0$ i.e. $z$ is a root of $F$ in $\Bbb C.$ Which proves $\Bbb C$ is algebraically closed.
I have proved that $F$ is a closed map on $\Bbb C$ when $\Bbb C$ is endowed with the usual metric topology. How do I prove that $F$ is an open map in order to prove the required result?
Any help will be highly appreciated. Thank you very much.

You use the open mapping theorem, which states that every analytic non-constant map from an open connected subset of $\mathbb C$ into $\mathbb C$ is open.
