I'd like to understand why if $X$ is a locally Noetherian scheme, then $X$ is quasiseparated. Recall that a scheme is quasiseparated if the intersection of two quasicompact open subsets is quasicompact. This is equivalent to the intersection of any two affine open subsets being a finite union of affine open subsets. Further, locally Noetherian means that $X$ can be covered by affine open sets $\operatorname{Spec}A$ where $A$ is a Noetherian ring.
So suppose that $U=\operatorname{Spec}A$ and $V=\operatorname{Spec}B$ are two open affine subsets of $X$. Further suppose that $X$ is covered by $\{\operatorname{Spec}A_i\}$, as $i$ runs over some index set $I$, and where the $A_i$ are Noetherian rings. We'd like to show that $U\cap V$ can be covered by a finite number of affine open subsets of $X$.
Recall that the intersection $U\cap V=\operatorname{Spec}A\cap\operatorname{Spec}B$ is a union of open sets that are simulaneously distinguished in both $\operatorname{Spec}A$ and $\operatorname{Spec}B$. Further, since $X$ is covered by $\{\operatorname{Spec}A_i\}_{i\in I}$, $U\cap V$ is also the union of open sets that are distinguished in some subset of $\{\operatorname{Spec}A_i\}_{i\in I}$. But how to show that this cover is finite? Or is there a better way to approach this?