[**Edit:** Question 1 has been moved elsewhere so that an answer to Question 2 can be *accept*ed.]

**Question 2.** *Is there a number field* $K$, *and a smooth proper scheme* $X\to\operatorname{Spec}(\mathfrak{o})$ *over its ring of integers, such that* $X(K_v)\neq\emptyset$ *for every place* $v$ *of* $K$, *and yet* $X(K)=\emptyset$ ?

I believe the answer is Yes.

**Remark.** Let $K$ be a real quadratic field, $\mathfrak{o}$ the ring of integers of $K$, and $A$ the quaternion algebra over $K$ which is ramified exactly at the two real places. Then the conic $C$ corresponding to $A$ is a smooth projective $\mathfrak{o}$-scheme such that $C(\mathfrak{o})=\emptyset$ (because $C(K_v)=\emptyset$ for each of the real places $v$). But if we insist that $C(K_v)\neq\emptyset$ at these two real places $v$, then $A$ would have to split at these $v$ (in addition to all the finite places), and we would have $C=\mathbb{P}_{1,\mathfrak{o}}$.

More generally, let $K$ be a number field, $\mathfrak{o}$ its ring of integers, and let $C$ be a smooth proper $\mathfrak{o}$-scheme whose generic fibre $C_{K}$ is a twisted $K$-form of the projective space of some dimension $n>0$. If $C$ has points everywhere locally, then $C=\mathbb{P}_{n,\mathfrak{o}}$. This remark shows that $X$ cannot be a twisted form of a projective space.