maybe you mean $s<\frac{1}{2}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=67555

jin wrote: maybe you mean $s<\frac{1}{2}$. Yes you're right !!! Thanks

Assume we can cover it with 5 squares. However look at a side of the unit square , since $s<\frac{1}{2}$ it must be covered by at least 3 of the 5 squares. Analogously the opposite side must be covered by at least 3 squares. Therefore at least one of these squares must cover both sides but this is impossible since its diagonal is less than $\frac{\sqrt{2}}{2}$ which is in turn less than 1. Therefore we have reached a contradiction and we are done. Correct me if I'm wrong please.

Yes, unfortunately you're wrong, since for $s<1/2$, but large enough, it is easy to use two squares to cover a side. A start towards a solution is to look at the nine points given by the vertices of the unit square, the midpoints of its sides, and its centre. No square of side $s<1/2$ may cover three of them, since if collinear they stretch on a length of at least $1 > s\sqrt{2}$, while otherwise they determine a triangle of area at least $1/8 > s^2/2$. Therefore by pigeonhole principle, we need at least five small squares.

Ok now I see that my mistake was assuming we purposefully had to place the squares inside the unit square. However your solution is not clear to me, you mention the fact that we can't cover those nine points simultaneously with the squares and then you prove that we can't do so with 4 squares? Since if you mean it's not possible with 5 I have an easy counterexample for that. The interesting thing about this problem is that it is so simple it should be possible to generalize. I'm sorry if I misunderstood your proof.