Smooth proper schemes over rings of integers with points everywhere locally [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.
 A: There is no doubt that such examples as in David Speyer's response exist: indeed, they exist in great abundance in the following sense:
Let $k_1$ be any number field, and let $E_{/k_1}$ be any elliptic curve with integral $j$-invariant.  Then it has potentially good reduction, meaning that there is a finite extension $k_2/k_1$ such that $E_{/k_2}$ is the generic fiber of an abelian scheme over
$\mathbb{Z}_{k_2}$.  Furthermore, let $N$ be your favorite integer which is greater than $1$. Then there exists a degree $N$ field extension $k_3/k_2$ such that the Shafarevich-Tate group of $E_{/k_3}$ has an element of order $N$ (in fact, one can arrange to have at least $M$ elements of order $N$ for your favorite positive integer $M$): see Theorem 3 of
http://alpha.math.uga.edu/~pete/ClarkSharif2009.pdf
Since good reduction is preserved by base extension, the genus one curve $C_{/k_3}$ corresponding to the locally trivial principal homogeneous space of $E_{/k_3}$ of period $N$ gives an affirmative answer to Question 2.
Specific examples of elliptic curves over quadratic fields with everywhere good reduction are known: see e.g. the survey paper
http://mathnet.kaist.ac.kr/pub/trend/shkwon.pdf
where the following example appears and is attributed to Tate:
$E: y^2 + xy + \epsilon^2 y = x^3, \ \epsilon = \frac{5+\sqrt{29}}{2}$,
has everywhere good reduction over $k = \mathbb{Q}(\sqrt{29})$.  Indeed, the given equation is smooth over $\mathbb{Z}_k$, since the discriminant is $-\epsilon^{10}$ and $\epsilon$ is a unit in $\mathbb{Z}_k$.
If this elliptic curve happens itself to have nontrivial Sha, great.  If not, the theoretical results above imply that a quadratic extension of it will have a nontrivial $2$-torsion element of Sha, i.e., there will exist some hyperelliptic quartic equation
$y^2 + p(x)y + q(x) = 0$
with $p(x), q(x)$ in the ring of integers of some quadratic extension $K$ of $\mathbb{Q}(\sqrt{29})$, which is smooth over $\mathbb{Z}_K$ and violates the local-global principle.
If someone is interested in actually computing the equation, I would say a better strategy is searching for elliptic curves defined over quadratic fields with everywhere good reduction until you find one which already has a 2-torsion element in its Shafarevich-Tate group.  (I don't see how to guarantee this theoretically, but I would be surprised if it were not possible.)  Then it is easy to write down the defining equation.
A: Regarding question 2, does the following work? Let $E$ be a rational elliptic curve with integer $j$-invariant. Then there is a number field $K$ so that $E \times_{\mathbb{Q}} K$ has a smooth model over $\mathcal{O}_K$. Roughly, $\mathbb{Q}(j^{1/6})$ should work, but there might be some subtleties at 2 and 3. If Sha of $E \times_{\mathbb{Q}} K$ is nontrivial, then I think an element of Sha should correspond to a torsor for $E \times_{\mathbb{Q}} K$ with the required property.
I don't understand the elliptic curve tables well enough to know how to search them for an example like this, but presumably one of our readers does.
A: Chandan asked Vladimir and me for an example of an elliptic curve over a real quadratic field that has everywhere good reduction and non-trivial sha, with an explicit genus $1$ curve representing some element of sha. Here's one we found:
The elliptic curve $y^2+xy+y = x^3+x^2-23x-44$ over $\mathbb Q$ (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0* over $K=\mathbb Q(\sqrt{65})$, and I0* can be killed by a quadratic twist. Specifically, the original curve can also be written as $y^2 = x^3+5x^2-360x-2800$ over $\mathbb Q$, and its quadratic twist over $K$.
$E: \sqrt{65}Uy^2 = x^3+5x^2-360x-2800$
has everywhere good reduction over $K$; here $U = 8+\sqrt{65}$ is the fundamental unit of $K$ of norm $-1$. 
Now 2-descent in Magma says that the 2-Selmer group of $E/K$ is $(\mathbb Z/2\mathbb Z)^4$, of which $(\mathbb Z/2\mathbb Z)^2$ is accounted by torsion. So it has either has rank over K or non-trivial Sha[2], and according to BSD its rank is 0 as L(E/K,1)<>0 (again in Magma). Actually, because $K$ is totally real, I think results like those of Bertolini and Darmon might prove that E has Mordell-Weil rank $0$ over $K$ unconditionally. So it has non-trivial Sha[2]. After some slightly painful minimisation, one of its non-trivial elements corresponds to a homogeneous space  
$C: y^2 = (23562U+1462)x^4 + (4960U+240)x^3 + (1124U-291)x^2 + (141U-833)x + (50U-733)$
with $U$ as above. So here is a curve such that $J(C)$ has everywhere good reduction and the Hasse principle fails for C.
Hope this helps!
Tim
A: If the fibres of the morphism $f: X\rightarrow\mathrm{Spec}(\mathbb{Z})$ have dimension 
$\leq 1$ the following facts are interesting regarding question 1:


*

*By a theorem of Minkowski the field $\mathbb{Q}$ has no unramified extensions.
If the fibres of $f$ have dimension $0$ smoothness is the same as being etale thus
leading to an unramified extension of $\mathbb{Q}$.

*A theorem of Fontaine proved in 1985 says that there exist no proper smooth
curves over $\mathbb{Q}$ of genus $g\geq 1$ with good reduction everywhere.
Thus the case of a non-rational curve of genus $0$ remains, that is a curve with
$g=0$ and without a rational point. 
