Let $(X,O_X)$ be the ringed space associated to a "classical projective" variety over an algebraically closed field $k$. (i.e. $O_X$ is sheaf of regular functions, in the sense of Chapter I. Hartshorne. or Chapter 5,6 this notes ). Is it true that $$O_X(X)\simeq \bigcap_{p \in X} O_{X,p}?$$ More generally, is it true that for any open
$$O_X(V)= \bigcap_{p\in V} O_{X,p} ?$$ (Is there an analogous characterization for ringed spaces induced from schemes?)
I could prove this in the case of affine varieties: this follows from that if $B$ is an integral domain, then $$B=\bigcap B_{m} \subset Frac(B)$$ where $m$ ranges all the maximal ideals.
Any references would be appreciated.
First, you need to assume $X$ is irreducible (which is maybe part of your definition of "variety") for the question to even make sense: what does $\bigcap_{p \in X} O_{X,p}$ even mean? When $X$ is irreducible, you can consider all of these local rings as subrings of the function field of $X$ and take their intersection there. Without this assumption, though, there is no natural way to consider all the local rings as subrings of some other ring.
In any case, here is a more general statement:
Theorem: Let $X$ be an integral scheme in which every nonempty closed set contains a closed point (this is automatic if $X$ is Noetherian, for instance). Then $O_X(X)=\bigcap_p O_{X,p}$ as subrings of the function field of $X$, where $p$ ranges over all closed points of $X$.
Proof: Let $K(X)$ denote the function field of $X$. An element $f\in K(X)$ is in $O_X(X)$ if for every $p\in X$, there is a neighborhood $U$ of $p$ such that $f\in O_X(U)$. Indeed, if this is the case, then these local sections in $O_X(U)$ can be glued together to give a global section of $O_X$. But now if $f\in O_{X,p}$, that by definition means that there is a neighborhood $U$ of $p$ such that $f\in O_X(U)$.
So, we have $O_X(X)=\bigcap_{p\in X}O_{X,p}$. All that remains is to reduce the intersection down to just the closed points. Now note that if $q\in X$ is any point, then there is a closed point $p$ in $\overline{\{q\}}$. Then $O_{X,p}\subseteq O_{X,q}$, since every neighborhood of $p$ is also a neighborhood of $q$. So, $\bigcap_{p\in X}O_{X,p}$ is the same as the intersection where you only consider closed points.