$\newcommand{\Spec}{\mathrm{Spec}}$
Let $R \rightarrow S$ be a faithfully flat (unital) homomorphism of commutative rings. Does it follow that the corresponding map of topological spaces $\Spec S \twoheadrightarrow \Spec R$ is open?
If not, is there a counterexample where both $R$ and $S$ are finitely generated over the same algebraically closed field?
EDIT: The second question was answered in a comment, but I'm still interested in the first part.