It's Vakil's FOAG Exercise 5.1.H. The full statement is:
5.1.H. EXERCISE. Show that a scheme $X$ is quasiseparated if and only if there exists a cover of $X$ by affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.
And the definition of quasiseparated is:
A topological space is quasiseparated if the intersection of any two quasicompact open subsets is quasicompact.
the ⇒direction of this exercise is easy to show, but the other direction seems much harder. I found a related thread Quasiseparated if finitely covered by affines in appropriate way (it adds the quasicompact condition, which is now Exercise 5.1.I in Vakil's FOAG) and mimic the proof in the answer (it's cumbersome, I ask for a solution verification here, but feel free to skip it and give your proof):
(Proof for ⇐) Now suppose $X = \cup_{\alpha \in H} \operatorname{Spec} A_\alpha$ and for any two $\operatorname{Spec} A_\alpha$ and $\operatorname{Spec} A_{\alpha^\prime}$, the intersection $\operatorname{Spec} A_\alpha \cap \operatorname{Spec} A_{\alpha^\prime} =\cup_{i \in I_{\alpha,\alpha^\prime}} \operatorname{Spec} B_i$, i.e., is covered by a finite number of affine open subsets $ \cup_{i \in I_{\alpha,\alpha^\prime}} \operatorname{Spec} B_i$ ("covered by" here should mean equality). Here $H$ is some index set (maybe not finite) and for all pairs $(\alpha, \alpha^\prime)$, the index set $I_{\alpha,\alpha^\prime}$ is finite.
Let $\mathcal{B}$ be the collection consisting of all distinguished open subsets of all $\operatorname{Spec} A_\alpha$, i.e., $D_i \in \mathcal{B}$ iff there is some $ \operatorname{Spec} A_\alpha$ such that $D_i$ is a distinguished open subset in $\operatorname{Spec} A_\alpha$. Then $\mathcal{B}$ is also a base for $X$: any open set is the union of all the intersections of itself with $\operatorname{Spec} A_\alpha$ and each intersection is the union of some elements in $\mathcal{B}$, for the distinguished open subsets forming a base.
And the intersection $\operatorname{Spec} A_\alpha \cap \operatorname{Spec} A_{\alpha^\prime} =\cup_{i \in I_{\alpha,\alpha^\prime}} \operatorname{Spec} B_i$ can be changed to $\operatorname{Spec} A_\alpha \cap \operatorname{Spec} A_{\alpha^\prime} =\cup_{i \in I_{\alpha,\alpha^\prime}} D_i$, where $D_i$ is some distinguished open subset in $\operatorname{Spec} A_\alpha $ or $\operatorname{Spec} A_{\alpha^\prime}$. And we can assume all $D_i$ are distinguished open subsets of $\operatorname{Spec} A_\alpha$ or $\operatorname{Spec} A_{\alpha^\prime}$ in the same time.
We prove that following statement first: for every quasicompact open subset $U$ of $X$ and all $\operatorname{Spec} A_\alpha$, the intersection $U \cap \operatorname{Spec} A_\alpha$ is quasicompact. (The idea follows from the link above):
Suppose $U = \cup_{\alpha \in H'} \cup_{j \in J_\alpha} D_j$, here $H'$ is a finite subindex of $H$ and all index sets $J_\alpha$ are finite (the finiteness comes from $U$ being a quasicompact subset), and $D_j \subset \operatorname{Spec} A_\alpha$ is a distinguished open subset of $\operatorname{Spec} A_\alpha$. Then
$$ \begin{aligned} U \cap \operatorname{Spec} A_\alpha &\stackrel{(1)}{=} (\cup_{\alpha^\prime \in H'} \cup_{j \in J_{\alpha^\prime}} D_j) \cap \operatorname{Spec} A_\alpha\\ &\stackrel{(2)}{=} \cup_{\alpha^\prime \in H'} \cup_{j \in J_{\alpha^\prime}} (\operatorname{Spec} A_\alpha \cap D_j) \\ &\stackrel{(3)}=\cup_{\alpha^\prime \in H'} \cup_{j \in J_{\alpha^\prime}}\cup_{i \in I_{\alpha,\alpha^\prime}} (D_i\cap D_j) \end{aligned}\tag{*} $$
In the above equation $(*)$, the equality $(1)$ is from $U = \cup_{\alpha^\prime \in H'} \cup_{j \in J_{\alpha^\prime}} D_j$ where we change the index $\alpha$ to $\alpha^\prime$ for clarity, and assume that $D_j$ is a distinguished open subset in $\operatorname{Spec} A_{\alpha^\prime}$. The equality $(3)$ is from $\operatorname{Spec} A_\alpha \cap \operatorname{Spec} A_{\alpha^\prime} =\cup_{i \in I_{\alpha,\alpha^\prime}} D_i$ and assumes that all $D_i$ is a distinguished open subset in $\operatorname{Spec} A_{\alpha^\prime}$, as mentioned above earlier.
Now $D_i \cap D_j$ is still a distinguished open subset in $\operatorname{Spec} A_{\alpha^\prime}$. Hence it’s the spectrum of some localization of $A_{\alpha^\prime}$ and hence quasicompact. Hence $U \cap \operatorname{Spec} A_\alpha$ is quasicompact for being finite union of quasicompact space.
Given any two quasicompact open subsets $U, V$ of $X$, $$U \cap V = \cup (U \cap V \cap \operatorname{Spec} A_\alpha) = \cup ((U \cap \operatorname{Spec} A_\alpha) \cap (V \cap \operatorname{Spec} A_\alpha))$$ Since $U$ and $V$ are both quasicompact, we can assume there is a finite subindex of $\operatorname{Spec} A_\alpha$ covering $U \cup V$. Then the above equation can be assumed as finite union of quasicompact space, and hence quasicompact. ∎
Is this proof correct? Is there any other proof of this? Thank you very much.