In a finite-type algebra over a field, closed points are dense in every closed subset of the spectrum

If $U\subseteq Z$ is a nonempty open subset, choose $\mathfrak{p}\in U$. By [L2], there exists $f\in A$ such that $\mathfrak{p}\in D(f)\cap Z\subseteq U$. By [L1], write $Z=V(I)$ for a radical ideal $I$. In the quotient $B=A\mathrm{/}I$, the class $\bar{f}$ is not nilpotent, because otherwise every prime of $B$ would contain $\bar{f}$, contradicting $\mathfrak{p}\in D(f)\cap Z$.

The quotient $B=A\mathrm{/}I$ is a finite-type $k$-algebra by [A1]. Apply [L4] in $B$ to the radical ideal $(0)$. Since $\bar{f}$ is not in the intersection of all maximal ideals of $B$, there exists a maximal ideal $\mathfrak{n}\subseteq B$ with $\bar{f}\mathrm{\notin }\mathfrak{n}$. Let $\mathfrak{m}\subseteq A$ be the preimage of $\mathfrak{n}$. Then $I\subseteq \mathfrak{m}$ and $f\mathrm{\notin }\mathfrak{m}$, so $\mathfrak{m}\in D(f)\cap Z\subseteq U$. By [L3], $\mathfrak{m}$ is a closed point of $\mathrm{Spec}(A)$.

Step 2.1 shows that every nonempty open subset of $Z$ contains a closed point. This is exactly the density of the closed points inside $Z$.