The answer is somewhat suprisingly simple – it's $\pi/6$, in agreement with Henry's numerical results.
As Mark pointed out, the approach suggested by Arthur leads to a complicated set of constraints bounding the area in which the second point may lie. A different approach is to parametrize the admissible pairs of points using the circles that they're on and integrate over the Jacobian of their Cartesian coordinates. Thus,
$$
\begin{align}
x_1&=x+r\cos\phi\;,\\
y_1&=y+r\sin\phi\;,\\
x_2&=x-r\cos\phi\;,\\
y_2&=y-r\sin\phi\;,
\end{align}
$$
where $r$ is the circle's radius, $x,y$ are the coordinates of its centre and $\phi$ is the orientation of the diameter. The Jacobian matrix is
$$
\frac{\partial(x_1,y_1,x_2,y_2)}{\partial(x,y,r,\phi)}=\pmatrix{1&0&1&0\\0&1&0&1\\\cos\phi&\sin\phi&-\cos\phi&-\sin\phi\\-r\sin\phi&r\cos\phi&r\sin\phi&-r\cos\phi}\;.
$$
The $2\times2$ matrices in the lower half are the Jacobian matrices of polar coordinates with opposite signs of $\phi$; their determinants are $r$, and the overall determinant is $4r$.
Now consider the octant $0\le y\le x\le1$ of the square $[-1,1]^2$. In this region, the radius is bounded by $1-x$. The measure of all pairs of points in the square is $4^2=16$, so the desired probability is
$$
\begin{align}
p
&=
\frac8{16}\int_0^1\mathrm dx\int_0^x\mathrm dy\int_0^{1-x}\mathrm dr\,4r\int_0^{2\pi}\mathrm d\phi
\\
&=
2\pi\int_0^1\mathrm dxx(1-x)^2
\\
&=
\frac\pi6\;.
\end{align}
$$
Well although I doesn't make sense to put the words prove and probability in the same sentence, I'll answer this anyway. The question should be rephrased as find the probability and not prove the probability.
Anyway, the simplest definition of probability is No. of favorable outcomes divided by Total NO. of outcomes. Since, we are selecting a point from the square, the area of the square is the total no of outcomes. And, the favorable is the area of the circle, since, that is where we want the point. So, answer is area of circle divided by area of square.
Best Answer
A unit square is 1 x 1. Any point within 2/22 of the edge of the square will overlap the square. So the side length of the square boundary within the 1 x 1 square would be $1 - 2/22 - 2/22$, which is $(18/22)^2$, because there is a boundary on all sides of the square. So the probability is $\frac{(18/22)^2}{1} = (18/22)^2$