Path connected coloured sets on the squared paper Colour small squares on the standard squared paper in two colors A, B. Name two small squares with common side as "neighbor".
Let every colored set be "path connected": for any two small squares of the color A(resp. B) there is a sequence of color A(resp. B) neighbor squares from one to another. 
Could you help me to prove that must there exists square $3\times 3$ which has 6 squares of same color? 
(It is clear that there is infinite path of each color... I constructed a lot of finite examples without desired square $3\times 3$, but they haven't common structure...)
Question: does this $3\times 3$ square exists? I think, yes, but I can't prove it.
Added: There is a counterexample, two spirals without desired square $3\times 3$.
 A: I think the result is false.  Consider a sequence of drawings, one of which I will
represent here:

&&&&&&&&&&&&&&&
&  &  &  &  &
&  &  &  &  &
&& && && && &&
&  &  &  &  &
&  &  &  &  &
&& && && && &&
&& && && && &&


This is a coloring of a 9 x 15 region which satisfies the conditions and has no
3x3 square with six unit squares of the same color. (unfortunately, there are some
rendering problems as I am not seeing how to control the line spacing.)
 It should be clear how to extend 
this for mxn regions 
in which both m and n are arbitrarily large.  Now the idea is to develop a compactness
style argument which expresses connectedness of both regions, the lack of a 3x3
subregion with at least 6 squares of one color, and the arbitrary size of the diagram.
While I do not have the argument nailed down, I suspect one can use this to show an
infinite domain colored in such a way as to preserve all the properties.  This (plus
other poster's evidence to the contrary) is why I believe the poster's assertion that
such a 3x3 square exists that contains at least 6 squares of one color is false.

Gerhard "Ask Me About System Design" Paseman, 2010.09.05
A: Edit: This argument is wrong, as pointed out in the comments. I'll leave it up as a warning to others.
I am going to respond to the question under the assumption that this is about infinite paper (as mentioned by the original poster in the comments), rather than large finite paper, as I think that makes the proof a bit easier. Write $\mathcal{B}$ for the black region and $\mathcal{W}$ for the white region. If either of the regions is bounded then the result is obvious, as in this case there will be either an all-white or an all-black 3x3 square sufficiently far from the origin. So assume neither region is bounded. 
Assuming both regions are unbounded, it must be the case (possibly after switching black and white and maybe also a rotation of the plane by $\pi/2$) that $\mathcal{B}$ contains a two-way infinite path through the origin that contains squares with arbitrarily large $x$-coordinate, and also arbitrarily small $x$-coordinate. Since $\mathcal{W}$ is connected, all white squares must lie to one side of this path; by a reflection we can assume they all lie above the path. Then the region below the path is all black and so contains the square you're looking for. 
