Can separated schemes separate points. This question is related to a comment in Shafarevich's Basic Algebraic Geometry 2. 
Shafarevich shows that the affine line with doubled origin is not separated by showing that the diagonal is not closed in the product $X\times_k X$. 
Moreover, he shows that a rational function which is regular at one origin is also regular at the other origin (which is equivalent to $\mathcal{O}_{X,x_0}=\mathcal{O}_{X,x_1}$).
Then he says and I quote "It can be shown that nonseparatedness is quite generally associated with this type of phenomenon". 
So what is the general statement here. Naively one/I would think that he was saying something like this
For an integral separated scheme $X$ and points $x,y\in X$ such that $x\neq y$, there exists $f,g\in K(X)$ such that $f\in \mathcal{O}_{X,x},g\in \mathcal{O}_{X,y}$ and $f\notin \mathcal{O}_{X,y},g\notin \mathcal{O}_{X,x}$. 
So what is it that Shafarevich was alluding to.
 A: a) In general it is impossible given two points $x,y\in X$ of a separated integral scheme   to find $f\in \mathcal{O}_{X,x} $ with $f\notin \mathcal{O}_{X,y}$.
For example if $y$ is the generic point of $X$ then every $f\in \mathcal O_{X,x}$ is in $ \mathcal O_{X,y}=K(X)$, since $\mathcal{O}_{X,x} \subset K(X).$ 
b) However if $X$ is quasi-projective over a field and $x\neq y\in X$ are both closed you can always find $f\in \mathcal{O}_{X,x} $ such that $f\notin \mathcal{O}_{X,y}$.
Indeed,  one can find an open affine subscheme $U=\operatorname {Spec}(A) \subset X$ such that $x,y\in U$  (see here).
Now   $x\neq y$ correspond to maximal ideals $\mathfrak m\neq\mathfrak n\subset A$ and since there exists $0\neq g\in A $ with $g\in \mathfrak n, g\notin\mathfrak m$ we have for for $f=\frac 1g$  the desired result:$f\in \mathcal{O}_{X,x}=A_\mathfrak m $ but $f\notin \mathcal{O}_{X,y}=A_\mathfrak n$.
Since Shafarevich only considers quasi-projective classical varieties all of whose points are closed, this might explain what he had in mind  
