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

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.

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

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.

This is a variation on Poonen's question, taking Buzzard's fabulous example into account :

Question 1. Is there a smooth proper scheme $X\to\operatorname{Spec}(\mathbb{Z})$ such that $X(\mathbb{Q}_v)\neq\emptyset$ for every place $v$ of $\mathbb{Q}$ (including $v=\infty)$, and yet $X(\mathbb{Z})=\emptyset$ ?

I believe the answer is No.

Question 2. Is there such an example over (the ring of integers of) some number field ?

I believe the answer is Yes.

Remark 1. 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 $C=\mathbb{P}_{n,\mathfrak{o}}$.

Remark 2. Colliot-Thélène and Xu give a systematic treatment of quasi-projective $\mathbb{Z}$-schemes $Y$ such that $Y(\mathbb{Z}_p)\neq\emptyset$ for every prime $p$ and $Y(\mathbb{R})\neq\emptyset$, but $Y(\mathbb{Z})=\emptyset$. Some of these schemes might even be smooth, but none of them is proper.

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

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.

This is a variation on Poonen's question, taking Buzzard's fabulous example into account :

Question 1. Is there a smooth proper scheme $X\to\operatorname{Spec}(\mathbb{Z})$ such that $X(\mathbb{Q}_v)\neq\emptyset$ for every place $v$ of $\mathbb{Q}$ (including $v=\infty)$, and yet $X(\mathbb{Z})=\emptyset$ ?

Question 2. Is there such an example over (the ring of integers of) some number field ?

Remark 1. 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 $C=\mathbb{P}_{n,\mathfrak{o}}$.

Remark 2. Colliot-Thélène and Xu give a systematic treatment of quasi-projective $\mathbb{Z}$-schemes $Y$ such that $Y(\mathbb{Z}_p)\neq\emptyset$ for every prime $p$ and $Y(\mathbb{R})\neq\emptyset$, but $Y(\mathbb{Z})=\emptyset$. Some of these schemes might even be smooth, but none of them is proper.

This is a variation on Poonen's question, taking Buzzard's fabulous example into account :

Question 1. Is there a smooth proper scheme $X\to\operatorname{Spec}(\mathbb{Z})$ such that $X(\mathbb{Q}_v)\neq\emptyset$ for every place $v$ of $\mathbb{Q}$ (including $v=\infty)$, and yet $X(\mathbb{Z})=\emptyset$ ?

I believe the answer is No.

Question 2. Is there such an example over (the ring of integers of) some number field ?

I believe the answer is Yes.

Remark 1. 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 $C=\mathbb{P}_{n,\mathfrak{o}}$.

Remark 2. Colliot-Thélène and Xu give a systematic treatment of quasi-projective $\mathbb{Z}$-schemes $Y$ such that $Y(\mathbb{Z}_p)\neq\emptyset$ for every prime $p$ and $Y(\mathbb{R})\neq\emptyset$, but $Y(\mathbb{Z})=\emptyset$. Some of these schemes might even be smooth, but none of them is proper.