Consider, for arguments sake, that the particular configuration is still squares.
We know that for the case of squares, it can be done with a finite number of colours. Call it X.
Then, consider a different configuration where just one of the offices is some finite number of multiples long compared to the rest, such that it intersects with another copy of itself.
This requires at least one more colour. We now need X plus 1 colours. (which is false in 2d, we don't need 1 more color)
But then we could stretch the same rectangle again, and need one more colour. Ad infinitum. (no, we will never…
