Closer to centre than to any side

Solution. Split the square into four triangles by its two diagonals. Consider the top triangle $T = \{(x,y) : y \geq |x|\}$: here the closest side is the top side $y = \tfrac12$, at distance $\tfrac12 - y$.

In $T$, the point $(x,y)$ is closer to $O = (0,0)$ than to the top side iff $$\sqrt{x^2 + y^2} \leq \tfrac12 - y.$$ Both sides are non-negative on $T$, so squaring is reversible: the condition becomes $x^2 + y^2 \leq \tfrac14 - y + y^2$, i.e., $y \leq \tfrac14 - x^2$. This is the region under a downward parabola with focus $O$ and directrix $y = \tfrac12$.

The parabola $y = \tfrac14 - x^2$ meets the boundary line $y = x$ of $T$ where $x = \tfrac14 - x^2$, giving $x^* = \tfrac{\sqrt 2 - 1}{2}$. On $T$, the region both inside $T$ (so $y \geq |x|$) and below the parabola is $$R_T = \{ (x,y) : |x| \leq x^*,\; |x| \leq y \leq \tfrac14 - x^2 \}.$$ Its area equals $$\operatorname{area}(R_T) = 2 \int_0^{x^*} \left( \tfrac14 - x^2 - x \right) dx = 2 \left[ \tfrac{x^*}{4} - \tfrac{(x^*)^3}{3} - \tfrac{(x^*)^2}{2} \right].$$ With $x^* = (\sqrt 2 - 1)/2$ we have $(x^*)^2 = (3 - 2\sqrt 2)/4$ and $(x^*)^3 = (5\sqrt 2 - 7)/8$. Substituting and simplifying, $$\operatorname{area}(R_T) = \frac{4\sqrt 2 - 5}{12}.$$ By the fourfold symmetry of the square, the full region where $P$ is closer to $O$ than to every side has area $$4 \cdot \frac{4\sqrt 2 - 5}{12} = \frac{4\sqrt 2 - 5}{3}.$$ Since the unit square has area $1$, $$\boxed{\mathbb{P} = \frac{4\sqrt 2 - 5}{3} \approx 0.219.} \qedhere$$