Let $V\subseteq U$ be open subsets of $X=\text{Spec } R$, where $R$ is a commutative ring. So $V$ is the set of prime ideals not containing some ideal $I$, and $U$ is the set of prime ideals not containing some ideal $J$. Then in particular, $J\subseteq I$. Then, given a prime ideal $P\in V$, and an $f$ such that $X_f$, which is the set of prime ideals not containing $f$, is contained in $U$, I want to find an $f'$ such that $P\in X_{f'}\subseteq V\cap X_f$.
But I'm not sure that such an $f'$ is guaranteed to exist: Since $V$ is the set of prime ideals not containing $J$, $J$ is the union of the principal open sets $X_j$ for $j\in J$. Thus, I only need to show that $P$ is in one of the intersections $X_j\cap X_f=X_{jf}$. If this is not the case, then P must contain $jf$ for all $j\in J$. Since $P$ does not contain $J$, this means that $P$ contains $f$. But I don't see how this can be a contradiction, since $U$, and thus $V$ may contain prime ideals that do have $f$ as elements.
Chose $g \in I\setminus P$ and let $f^{\prime}=fg$