I'm focusing on the language of classic algebraic geometry.
Let $X$ be an affine variety and $A(X)$ be its coordinate ring. Let $E,F\subseteq X$ be $2$ disjoint closed subsets of $X$. Does there exist an $f\in A(X)$ such that $f(E) = 0$ and $f(F) = 1$?
If this is true, is there any generalization of this or in the language of scheme?
On
Let $I(E)$ and $I(F)$ be the regular functions vanishing on $E$ and $F$,
$I(E)$ is an ideal of $A(X)$ so its image in $A(X)/I(F)$ is an ideal.
If it is smaller than the whole ring $A(X)/I(F)$ then it must be contained in some maximal ideal $m$ containing both $I(F)$ and $I(E)$, which is impossible because $F$ and $E$ are disjoint.
Thus, there is some $f\in I(E)$ such that $f=1\in A(X)/I(f)$ ie. $f-1\in I(F)$.
A consequence is that $$A(X)/I(E)\times A(X)/I(F)\cong A(X)/(I(E\cup F)), \qquad (g,h)\to (1-f)g+fh$$
Yes, this is true. Let $R,S$ be the coordinate rings of $E,F$ respectively. Then $R\times S$ is the coordinate ring of $E\cup F$ as $E$ and $F$ are disjoint, and $A(X)\to R\times S$ is a surjection because $E\cup F$ is a closed subvariety/subscheme of $X$. So there is an element of $A(X)$ which has image $(0,1)\in R\times S$ and does exactly what you want. This proof works just fine for both affine schemes and affine varieties.