Let $X$ be a scheme which contains a closed point and also assume that for every two closed subsets $Y_1$ and $Y_2$ of $X$, we have either $Y_1 \subseteq Y_2$ or $Y_2 \subseteq Y_1$. Also assume that $X$ has finite Krull dimension. Then, is it true that $X$ is affine ?
My thoughts: Since the closed subsets of $X$ are comparable, so $X$ has exactly one closed point, say $x \in X$ and also every non-empty closed subset contains $x$ . Hence for any abelian sheaf $\mathcal F$ on $X$, we have $\mathcal F_x =\mathcal F(X)$. Since taking stalks of sheaves is exact, we get $H^j(X, \mathcal F)=0, \forall j>0$ for any abelian sheaf $\mathcal F$ on $X$. However , since I'm not assuming the scheme to be quasi-compact , I cannot quite apply Serre's criteria now.