In 2-by-2 partitions of two rectangles, is there always a pair of non-overlapping parts? There are two rectangles, red and blue.
Each rectangle is partitioned into 4 sub-rectangles using one cut parallel to its longer side and two cuts parallel to its shorter side, like this:

Does there always exits a pair of sub-rectangles, one red and one blue, that do not overlap?
NOTE: Initially I thought that the answer should be positive even when partitioning to 3 sub-rectangles. However, I found a negative example. In the partitions below, each blue sub-rectangle overlaps all three red sub-rectangles (and vice versa):

EDIT: here is a Geogebra worksheet to play with.
 A: I believe the answer is yes, there always exists a pair of sub-rectangles that do not overlap. Here is an outline proof.
Let the segment of the long cut that lies between the two points of intersection with the shorter cuts be called the Central Segment (CS).
Suppose that all four blue sub-rectangles overlap all four red sub-rectangles.
Then consider one blue sub-rectangle R. Each of the four red sub-rectangles must contain a point that is within R. Take four such points, one from each red sub-rectangle, and join them to make a quadrilateral Q.
Each edge of Q must lie within R (since it connects two points in R). Therefore Q lies fully within R.
Now Q must either intersect with the CS of the red rectangle, or it must fully enclose it (since the only way to draw such a quadrilateral without intersecting the red CS is to encircle it). Therefore the red CS passes through R.
The same is true for each blue sub-rectangle. Therefore the red CS passes through each blue sub-rectangle.
Let K and M be the two ends of the red CS, as shown in this diagram:

Note any segment that passes through each blue sub-rectangle - as the red CS does here - must cross both the blue CS and the two shorter cuts of the blue rectangle. (Such a segment may or may not continue beyond the bounds of the blue rectangle.)
Now consider the red sub-rectangle with a vertex at M whose edges do not overlap the red CS. There are two possible positions of this sub-rectangle, shown by the two dashed rectangles MOPL and MNQL in the diagram. But neither of these can possibly overlap with the blue rectangle that either contains or is nearest to K (the other end of the red CS), shown as GDFH in the diagram. This can be proved by (for example) considering the projection of the rectangle GDFH onto the line KL: this projection is entirely within the segment KM, while the projections of the potential red sub-rectangles MOPL and MNQL are the segment ML.
Therefore we have a red sub-rectangle that doesn't overlap with a blue sub-rectangle, and have reached a contradiction.
