Point inside a square
A point in the shaded areas above is closer to a side than a diagonal, with $AO$ and $BO$ bisecting $\angle CAB$ and $\angle ABC$ respectively, and $\triangle BOA$ repeated around the square because of symmetry. $O$ happens to be the incentre of $\triangle ABC$, and the inradius is
$$r=\frac{A(\triangle ABC)}s=\frac{1/4}{(\sqrt2+1)/2}=\frac{\sqrt2-1}2$$
The ratio $\frac{A(\triangle BOA)}{A(\triangle ABC)}=\frac{OD}{CD}$ then gives the desired probability:
$$\frac{OD}{CD}=\frac r{1/2}=\sqrt2-1=0.414213\dots$$