Let's take a scheme $X$. Is it possible to have an open non-empty subset $U$ of $X$ such that $\mathcal{O}_X(U)=0$?
I can't find an argument against it, since there could exist some open set $V\subset U$ such that $\mathcal{O}_X(V)\ne 0$, or is it impossible too?
It's obvious if $U$ were affine, but in the general case?
No, it is not possible: every non empty open subset $U\subset X$ has non zero ring of functions: $\mathcal{O}_X(U)\neq 0$ .
Indeed, if $V\subset U$ is a non empty open affine subset we get a restriction morphism of rings $r_{VU}:\mathcal{O}_X(U)\to \mathcal{O}_X(V)$ which is unity-preserving, like all ring morphisms.
Hence $r_{VU}(1_U)=1_V$ and since $\mathcal O_X(V )$ is the ring of functions of the non-empty affine scheme $V$ we have $1_V\neq0$, which forces $1_U\neq0$ and thus shows that $\mathcal{O}_X(U)\neq 0$.