Is the lower bound proof in this paper correct?
I get the same answer as you, $\frac{1+\sqrt{2}}{\sqrt{\pi}}$.

If I place two unit disks inside a larger square, their centers are one unit away from the left and right sides, and $\sqrt{2}$ units apart from each other in $x$-coordinates, so the side of the larger square is $2+\sqrt 2$. The unit disks have area $\pi$, and we want area $\frac{1}{2}$, so dividing the side length and the disk radius by $\sqrt{2\pi}$ gives the result.