Simplified proof of Chevalley's theorem for $A\to k$.

Question

I am looking for a simplier proof for the following special case of Chevalley's theorem:

Theorem. If $A\subset k$ is a subring of a field and the inclusion $i: A\to k$ is finitely presented, then $\{(0)\}$ is open in $\operatorname{Spec}(A).$ In other words, there is an element $f\neq 0$ contained in all non-zero prime ideals $\mathfrak p\subset A$.

See here for the proof of Chevalley's theorem in general for constructible sets. Are we able to reduce the proof significantly in this case?