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?
Let $U=\mathrm{Spec}(A) \subset X$ be an open of a locally Noetherian scheme $X$. Suppose $U_i=\mathrm{Spec}(A_i)$ is an open covering of $X$ by spectra of Noetherian rings.
As you said, $U \cap U_i$ can be cover by open that are distinguished (or principal) in both $U$ and $U_i$. In $U_i$, a distinguished open is Noetherian, being $\mathrm{Spec}((A_i)_g)$ for some $g \in A_i$. Thus $\mathrm{Spec}(A)$ is cover by (finitely many) distinguished opens $\mathrm{Spec}(A_f)$ with $A_f$ Noetherian. This implies that $A$ is Noetherian (maybe take this as an exercice). But the spectrum of a Noetherian ring is a Noetherian topological space; hence every open of $\mathrm{Spec}(A)$ is quasi-compact. In particular for another affine open $V$ of $X$, $U \cap V$ is quasicompact since it is an open of $U$