It is well-known in algebraic geometry that if $X$ is affine and $\mathcal{F}$ is a quasi-coherent sheaf on $X$, then $$ H^i(X,\mathcal{F})=0,~ \forall ~i\geq 1. $$
Now let $X$ be an arbitrary scheme and $\mathcal{F}$ a quasi-coherent sheaf on $X$. Let $U$ and $V$ be two affine open subsets of $X$, then we know that $$ H^i(U,\mathcal{F}|_{U})=0 \text{ and } H^i(V,\mathcal{F}|_{V})=0,~ \forall ~i\geq 1. $$
On the other hand, $U\cap V$ is no longer affine in general (if $X$ is not separated). Then do we still have $$ H^i(U\cap V,\mathcal{F}|_{U\cap V})=0,~ \forall ~i\geq 1 $$ or do we have counter-examples?
We may assume $X=U \cup V$. The Mayer-Vietoris sequence gives an exact sequence $$H^i(U,F) \oplus H^i(V,F) \to H^i(U \cap V,F) \to H^{i+1}(X,F) \to H^{i+1}(U,F) \oplus H^{i+1}(V,F)$$ which simplifies to $H^i(U \cap V,F) \xrightarrow{\,\cong\,} H^{i+1}(X,F)$. Since $H^{i+1}(X,F) \neq 0$ may happen when $i+1 \leq \dim(X)$, the answer is probably "no". (But I still have to find an explicit example ... $X$ has to be covered by two open affines, so that $\mathbb{P}^2$ won't work)