Non-projective smooth complete threefolds with a pair of points intersecting every surface I've been learning about non-projective complete varieties and I am trying to get a handle on how crazy they can get. 
$\textbf{Question:}$ Let $V$ be a complete threefold over $\mathbb{C}$. Given distinct points $P,Q$ on $V$, does there exist a complete surface $S\subset V$ which contains neither $P$ nor $Q$?
I would prefer it if $V$ were smooth, though I'd be surprised if that were relevant. I would also be interested in examples/proofs over fields of finite characteristic.
Thanks!
 A: I do not know the answer to the question. However, let me point out that the answer is negative if you look for surfaces avoiding more points. More precisely, there exists a smooth complete threefold that does not satisfy Sándor's property $NC_{10}$: it contains 10 points that cannot be simultaneously avoided by a divisor.
In [Fujino - Payne, Smooth complete toric threefolds with no nontrivial nef line bundles], the authors construct varieties as indicated in the title. Their best example $X$ has Picard rank $5$, and they show that it is optimal.
  Now, the property that $X$ has no nontrivial nef line bundle is equivalent to the fact that the cone of effective curves is the whole vector space $N_1(X)$ of $1$-cycles up to numerical equivalence. Since this vector space is $5$-dimensional, it will be possible to find $10$ integral curves $C_1,\dots,C_{10}$ on $X$ spanning this vector space as a cone (see [Davis, Theory of positive linear dependance, Theorem 6.7]). Choose a point $P_i$ in $C_i$. I claim that no surface in $X$ avoids all the $P_i$. Indeed, such a surface $S$ would intersect non-negatively the $C_i$, hence any curve. This is obviously impossible: working in an affine neighbourhood of a point of $S$, one constructs a curve intersecting $S$ positively.
Note that in Fujino and Payne's precise example, it should be possible to find less than $10$ integral curves spanning $N_1(X)$ as a cone, but certainly not less than $6$ curves.
Finally, toric varieties are never going to provide counter-examples to your original question, as any two points in a toric variety have a common affine neighbourhood.
A: This is just a partial answer to restrict the search.
Condition $NC_q$ Let us say that $Z$ satisfies condition $NC_q$ if for any $z_1,\dots,z_q\in Z$ there exists $D=D_{z_1,\dots,z_q}\subset Z$, a codimension one subscheme of $Z$ such that $z_i\not\in D$, for any $i=1,\dots,q$.
Example If $Z$ is projective, then it saisfies $NC_q$ for any $q\in\mathbb N$.

Claim
  Suppose that $Z$ is a scheme that satisfies $NC_q$ and let $f:V\to Z$ be a dominant morphism. Then $V$ satisfies $NC_q$.

Proof Apply the condition for $f(v_1),\dots,f(v_q)$ and take $f^{-1}(D)\subset V$.
Corollary If $V$ is an arbitrary threefold that admits a dominant morphism to a projective variety of positive dimension, then it satisfies $NC_2$. In particular, Hironaka's example of a non-singular non-projective complete threefold does not provide a counter-example.

There are non-projective schemes with trivial Picard group and hence without non-trivial line bundles. Obviously these cannot admit a non-trivial morphism to a projective space (variety). 
On the other hand, such a scheme has to be singular: Take an arbitrary dense affine subset and take an effective divisor there. The closure of this in the original scheme will give a non-trivial divisor which would then have to be a non-Cartier divisor, as otherwise the Picard group could not be trivial. But then the ambient scheme has to be singular.
Alternatively one could argue, using Chow's lemma that there are non-trivial divisors on any complete variety and hence the Picard group of a complete non-singular variety cannot be trivial. 
So, an example of a complete non-singular threefold failing $NC_q$ would have the property that every linear system has base points. I don't know if that exists.
For a singular example you could do the following (this is motivated by Jason's comment): Take a projective threefold whose Picard group is $\mathbb Z$ and hence every effective divisor is ample on it. Contract some curves to get something non-projective. Now take any surface on the target. The pre-image of that is an effective divisor on the original threefold which then has to be ample, and hence it has to intersect the contracted curves. Therefore the chosen surface has to contain the image(s) of these curves, so if you choose your $P,Q$ among these image points, then you get a counterexample.
