Distinguished opens cover $\Spec R$ exactly when their defining functions generate $1$; a finite subcover is already hidden in that equation.Suggest a slogan
Let \(R\) be a ring and \(\{f_i\}_{i \in I}\) elements of \(R\). Then \(\bigcup_i D(f_i) = \Spec R\) if and only if the \(f_i\) generate the unit ideal. In that case finitely many of the \(f_i\) already generate it, so \(\Spec R\) is quasicompact.
Show that $\{f_i\}$ generate the unit ideal of $R$ if and only if the $D(f_i)$ cover $\Spec R$.