let $X : Ring \to Set$ be a functor (a Zfunctor in the language of demazure, gabriel) and $V \subseteq X$ a locally closed subfunctor. assume that $U \subseteq V$ is an open subfunctor. does then exist an open subfunctor $W \subseteq X$ such that $U = V \cap W$? if $X$ and $V$ are schemes, this should be true.
I don't think this is true even when $X$ and $V$ are schemes if you only require that the map $V\to X$ is an embedding of functors (rather than a locally closed embedding). Example: Take $X=Spec(k[x])$, $V=Spec(k[x,x^{1}]\times k)$, $U=Spec(k)$. That is, $X$ is an affine line, $V$ is the disjoint union of a punctured line and origin, and $U$ is a point, which we view as the connected component of $V$. The natural map $V\to X$ (corresponding to stratification of $X$) is a categorical monomorphism: it induces an embedding of functors. EDIT: Now we add the assumption that $V\hookrightarrow X$ is locally closed. I think the statement still fails, here's a counterexample: $X$ will be an indscheme: so there is a sequence of schemes $X_n$ related by closed embeddings $X_n\hookrightarrow X_{n+1}$ and $$X(A)=\lim_{\to} X_n(A).$$ In other words, every point of $X(A)$ factors through one of $X_n$'s. For indschemes, a locally closed subfunctor $V\subset X$ is given by a compatible family of locally closed $V_n\subset X_n$ (so that it is a locally closed subindscheme). Problem is, the statement fails for indschemes. Let's take $X_n={\mathbb A}^n$, with the embedding $X_n\hookrightarrow X_{n+1}$ being the coordinate embedding. Now let $V_n$ be the union of the origin $0$ and $n1$ punctured lines $$l_k:=\lbrace(k,0,\dots,0,x,0,\dots,0)x\ne 0\rbrace,$$ where $x$ is in the $k$th position, and $k$ varies from $2$ to $n$. (Let's say I work over a field of characteristic $0$, so all integers are distinct.) Finally, $U_n\subset V_n$ is the origin, which is a component of each $V_n$, so it is both open and closed. It is easy to see that it is impossible to find a compatible family of open subsets $W_n\subset X_n$ such that $U_n=V_n\cap W_n$. Indeed, $W_1$ contains $0$, so it is nonempty, and thus contain $n\in{\mathbb A}^1=X_1$ for some $n$. But then $W_n$ must contain $(n,0,\dots,0)$ without meeting $l_n$. 

