TITLE: Do there exist two points seeing one another?
QUESTION [11 upvotes]: Let $n\ge1$ be an integer number. Let $n$ nonoverlapping closed line segments and 
$n+2$ distinct points which do not belong to those line segments be given in the plane 
$\mathbb{R}^2$.
Can two points among the $n+2$ given points be chosen such that the ones
 see one another i.e. the closed line segment connecting these points does not intersect
 any of the given $n$ line segments? In the case $n=1$ the affirmative answer is simple.
However, I don't know the answer even for $n=2$.
Addition. Here is a modification of Joseph picture which demonstrates the difficulties when realizing the idea by Per

If the points $A(\frac {38} {20},\frac {38} {20}) $ and $B(\frac {32} {10},\frac {32} {10} ) $ are assumed to belong to $\{(x,y):y \ge x,y\ge 3-x\} $,
 then the proof suggested by Per fails.
Addition 2.
I'd like to demonsrate another difficulty which appears in the Per's approach. Let us consider
three segments $S_1:=[(-1,-1),(1,-1)],\,S_2:=[(1,1),(1,2)],\,S_3:=[(-1,0),(2,1)]$. The points
are not of importance here. After the first step the plane is divided into two regions $R_1$ and $R_2$. 

After the second step we obtain
the regions $R_1$ and $R_3$, where the latter is not convex.

It is unclear for me whether after the final step we obtain the convex regions only (the larger the value of $n$, the more complicated the situation).
One more problem consists in the definitions of the borders of the regions. In order
to apply the pigeonhole principle the regions considered with their borders must not intersect.
My advice to Per is to read the book I. Lakatos. Proofs and Refutations. Cambridge: Cambridge University Press.(1976). 
https://books.google.com.ua/books/about/Proofs_and_Refutations.html?id=1n6SFdXCOBQC&redir_esc=y

REPLY [7 votes]: This very problem was proposed to St. Petersburg olympiad (selection round) in 2007 by Konstantin Kokhas (problem 6 for 10-th grade in the linked pdf). In the same year it was proposed - independently, I guess, - to the journal Matematicheskoe Prosveschenie (problem 5 in the linked pdf) by Maxim Kontsevich himself. 
Here is a solution which is hopefully self-contained and complete.
Assume the contrary. At first, we slightly enlarge the segments so that new segments still do not overlap. After that each segment $AB$ between our $n+2$ points intersects one of $n$ segments in an interior point. This property is preserved under small perturbation of the $n$ segments. Such perturbation allows to get $n$ segments such that no three lines containing these segments have a common point, and no three out of $2n$ their endpoints lie on a line (for example, you may choose new endpoints one by one so that each time you fix an endpoint it does not lie on a line between two already fixed endpoints; and each time you fix both endpoints of a segment, the line containing it does not pass through a common point of other two lines. This is clearly possible, since already finitely many lines are forbidden and a small disc is allowed.) 
Now we enlarge the segments one by one, each time either to infinity or to the intersection with another segment. We get a plane partitioned onto several convex regions. Let's prove that there are exactly $n+1$ regions and they are convex. Draw a large square and remove the parts of rays outside the square. Then we get a planar graph with all degrees equal to 3, and the number of vertices equals $2n$ (2 vertices at endpoints of each of extended segments). Each region is a convex polygon (since going along the boundary of each of the regions we see that all angles are less than $\pi$). Thus the number of edges equals $3n$, and the number of faces equals $n+2$ by Euler formula. One face is external, so $n+1$ inner faces as desired.