I wonder if the non-Noetherian scheme $\operatorname{Spec}k[x_1,x_2,x_3,\dots]$ is a locally Noetherian scheme.
I have no idea how to find a Noetherian affine open cover, if exists or prove that it is impossible to find a Noetherian affine open cover.
Let $X$ be a scheme. We say $X$ is locally Noetherian if every $x\in X$ has an affine open neighborhood $\text{Spec}(A)$ where $A$ is a Noetherian ring. It turns out that this is equivalent to the statement that for every affine open $U\subset X$, the ring $\mathcal{O}_X(U)$ is Noetherian. To show the equivalence, you want to show that being Noetherian is a property that is "affine local." See here: https://stacks.math.columbia.edu/tag/01OW. For your case, we apply the above equivalence, taking $U$ to be $X=\operatorname{Spec}k[x_1,x_2,\dots]$. But $k[x_1,x_2,\dots]$ isn't Noetherian, so $X$ can't be locally Noetherian.