Closer to the centre than to the edge

Idea. Split the square along its diagonals. In each of the four triangles the nearest side is fixed, and “closer to the centre than to that side” is the inside of a parabola. Integrate in one triangle and multiply by four.

Step 1: set-up. Put the centre at the origin, so the square is $[-\frac{1}{2},\frac{1}{2}{]}^2$. In the bottom triangle $\{y\le -|x|\}$ the nearest side is $y=-\frac{1}{2}$, at distance $y+\frac{1}{2}$. The point is closer to the centre when $$x^2+y^2<{\left(y+\frac{1}{2}\right)}^2⟺y>x^2-\frac{1}{4}.$$

Step 2: the limits. In the triangle we also need $y\le -|x|$. Both hold when $x^2-\frac{1}{4}<-|x|$, that is $|x|<a$ with $$a=\frac{\sqrt{2}-1}{2}.$$

Step 3: the area in one triangle. $$2{\int }_0^a\left(-x-x^2+\frac{1}{4}\right)dx=2\left(\frac{a}{4}-\frac{a^2}{2}-\frac{a^3}{3}\right).$$

Step 4: all four triangles. The total area is $2a-4a^2-\frac{8}{3}{a}^3$. With $2a=\sqrt{2}-1$, $4a^2=3-2\sqrt{2}$ and $8a^3=5\sqrt{2}-7$: $$\begin{array}{rl}& (\sqrt{2}-1)-(3-2\sqrt{2})-\frac{1}{3}(5\sqrt{2}-7)\\ & =\frac{4\sqrt{2}-5}{3}\approx \mathrm{0.219.}\end{array}$$

The trap. Guessing the inner square of side $\frac{1}{2}$, with area $\frac{1}{4}$. The true region bulges towards the corners along the diagonals and is pinched in the middle of each side, and its area is a little smaller.