Denote $ABCD$ the square and $S_1$, $S_2$, $S_3$ the discs.

Suppose that $S_1$, $S_2$ and $S_3$ cover the whole square. Since there are three discs and the square has four vertices, one of the discs must cover two vertices of the square, say $S_1$ covers $A$, $B$. Then $[AB]$ is a diameter for $S_1$, so $S_1$ cannot cover any point from $(BC] \cup [CD] \cup [DA)$. So, $C$ must be covered by one of the other discs, say $S_2$. Then $S_2$ cannot cover any point from $(AD)$, therefore $(AD) \subset S_3$. In this case $[AD]$ is a diameter of $S_3$, so $S_3$ cannot cover any point from $(BC)$, therefore $S_2$ must cover $(BC)$. This shows that $[BC]$ must be a diameter of $S_2$. But, in this case, no point from $(CD)$ is covered – contradiction.

We take as $S_1$, $S_2$, $S_3$ the discs of diameters $[AM]$, $[PT]$, $[RS]$. Denote $X$ the point on $[AC]$ for which $XT \perp BC$ (and $XU \perp CD$). Then $S_1$ covers the square surface $APMR$, $S_2$ covers the pentagonal surface $BPMXT$ and $S_3$ covers the pentagonal surface $DRMXU$, so the points not covered by the three discs are inside the square $CUXT$.

It is enough to prove that
$$\text{area}[CUXT] < 0.25\% \cdot \text{area}[ABCD],$$
which is equivalent to $CT < BC/20 = 0.1$, or $BT > 1.9$. Simple considerations yield $AP = \sqrt{2}$, $BP = 2 - \sqrt{2}$, $BT^2 = 4 - (2 - \sqrt{2})^2 = 4\sqrt{2} - 2$. We are left to prove $BT > 1.9$, that is $4\sqrt{2} - 2 > 1.9^2$, or $\sqrt{2} > 1.4025$, which is true.