Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? I was woolgathering about the notion of a scheme, and it occurred to me that I know of no non-affine scheme $S$ that is the union of $Spec(O_K)$'s of some number field $K$ (I allow $K$ to vary - so that $S$ might be $Spec(O_K)\cup Spec(O_L)$ for example).
It would an interesting notion if one could patch rings of integers together to form some non-affine $1$-dimensional normal scheme $S$. The fact that I've never seen an example makes me think it's impossible.
Question
Is there a connected non-affine scheme $S$ such that it is the union of open subschemes of it that are $Spec$'s of rings of integers of number fields?
More pointedly, if $Spec(O_K)$ (the ring of integers of some number field $K$) is an open subscheme of a normal scheme $S$ then is it equal to it?
 A: I say no.
Let $\xi_K$ be the generic point in $Spec(O_K)$, and $\xi_L$ the generic point in $Spec(O_L)$. Since $Spec(O_K)$ and $Spec(O_L)$ have nonempty intersection, their intersection must be an open set in each, and must contain both generic points. The local ring at $\xi_K$ is $K$, and the local ring at $\xi_L$ is $L$.But no point in $O_K$ has local ring $L$, and no point in $O_L$ has local ring $K$. This is a contradiction.
A: Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
 Take two copies $U'\subset X',U''\subset X''$ of the above   and glue them along the identity  $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
 The scheme  $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated. 
Edit I wasn't too happy with this non-separated example when I posted it, but Qing now has proved that it is impossible to find a separated one.
A: If $i: \mathrm{Spec}(O_K)\to S$ is an open immersion into a connected separate scheme $S$, then $i$ is an isomorphism. Indeed, the canonical morphism $\pi : \mathrm{Spec}(O_K)\to \mathrm{Spec}(\mathbb Z)$ is finite (hence proper) and can be decomposed into $i$ followed by the canonical morphism $S\to \mathrm{Spec}(\mathbb Z)$. As the latter is separated, this implies that $i$ is also proper, hence closed. The connectedness of $S$ implies that $i$ is onto. 
