The prime spectrum is compact in the library's non-Hausdorff sense

If $\mathrm{Spec}(R)=\mathrm{\varnothing }$, then the empty subfamily of $\mathcal{U}$ already covers it. By [L1], the spectrum is compact in this case.

Assume now that $\mathrm{Spec}(R)\mathrm{\ne }\mathrm{\varnothing }$. For each $\mathfrak{p}\in \mathrm{Spec}(R)$, choose $U_{\mathfrak{p}}\in \mathcal{U}$ with $\mathfrak{p}\in U_{\mathfrak{p}}$. By [L2], choose $f_{\mathfrak{p}}\in R$ with $\mathfrak{p}\in D(f_{\mathfrak{p}})\subseteq U_{\mathfrak{p}}$. Then $\{D(f_{\mathfrak{p}}){\}}_{\mathfrak{p}\in \mathrm{Spec}(R)}$ is a distinguished-open cover of the spectrum.

By [L3], the ideal generated by the family $\{f_{\mathfrak{p}}\}$ is $R$. Hence there exist finitely many primes ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n\in \mathrm{Spec}(R)$ and coefficients $a_1,\dots ,a_n\in R$ such that $1=a_1{f}_{{\mathfrak{p}}_1}+\cdots +a_n{f}_{{\mathfrak{p}}_n}$.

Applying [L4] to this identity gives $\mathrm{Spec}(R)=D(f_{{\mathfrak{p}}_1})\cup \cdots \cup D(f_{{\mathfrak{p}}_n})\subseteq U_{{\mathfrak{p}}_1}\cup \cdots \cup U_{{\mathfrak{p}}_n}$. Thus $\mathcal{U}$ has a finite subcover.

Steps 1.1 and 3.1 show that every open cover of $\mathrm{Spec}(R)$ has a finite subcover. Therefore $\mathrm{Spec}(R)$ is compact by [L1].