[GeeTwo]

After seeing the original post earlier today, I did not check back until I had this. It does not seem that anyone got as far as I did:

Initial thoughts and reasoning:
After looking at the problem for about a minute, my "eyeball integrator" came up with "a bit ovor 0.5". While I did not go through all these steps conciously, this is roughly what I think happened:

Consider a circles of radius 1/2 unit and 1 unit. Considering 25 points in a 5x5 grid, including the corners of the square and the center of the square.
When the circles are co-centered with the square, about 79% of the square is inside the small circle, all within the large circle. There is one such point.
When the circles are located halfway between the center and an edge, the small circle covers 66% of the square, the large circle all of the square. There are four such points.
When the circles are located halfway between the center and a corner, the small circle covers 55% of the circle. The large circle covers 99+% of the square (ignore). There are four such points.
When the circles are located at the center of an edge, the small circle covers 37% of the large square. About 4% of the square is outside of the large circle. There are four such points, but only half of a square centered here is inside the square. Weight=2.
When the circles are located halfway between the previous point and a corner, the small circle covers 33% of the square. About 8% of the square is outside the large circle. There are 8 such points, only half of the square centered here is inside the big square. Weight = 4.
When the circles are located at the corner, only 18% of the square is inside the small circle, and 21% of the square is outside the large circle. There are four points, onle a quarter of a square centered here is inside the big square. Weight=1.
Therefore, for an arbitrary point, .79 + .18 + 2 * .37 + 4 * (.66 + .55 + .33) / 16 = 49% of segments are shorter than 1/2 a unit. Likewise, (.04*2 + .08 * 4 + .21 * 1)/16 = 4% of the segments are longer than a full unit. I expect the answer to be close to a half unit, perhaps a bit larger.

Next, a numerical solution:

#!/bin/gawk -f
BEGIN {
  for (steps=10; steps<200; steps+=10){
    numer=denom=0;
    for (i=0; i< steps; i++)
      for (j=0; j< steps; j++)
        for (k=0; k< steps; k++)
          for (l=0; l< steps; l++) {
            numer += sqrt((i-j)*(i-j)+(k-l)*(k-l));
            denom+=steps;
    }
    print steps, numer/denom
  }
}

Output: [code]
10 0.518687
20 0.520757
30 0.521121
40 0.521247
50 0.521304
60 0.521336
70 0.521354
80 0.521366
90 0.521375
100 0.52138
110 0.521385
120 0.521388
130 0.521391
140 0.521393
150 0.521394
160 0.521396
170 0.521397
180 0.521398
190 0.521399

This appears to meet Ether's criterion of 5 significant digits at ".52140".

Nonetheless, let's try to find a closed-form solution. As indicated in my numerical solution, the problem is at its most basic the ratio of two quadruple integrals over the interval of [0,1], the content of the numerator integral being the hypotenuse formula and the denominator being unity. As integrating 1 over the range 0 to 1 yields 1, doing this four times still yields one, so we can skip the denominator. I recognize that this solution counts all of the non-zero-length segments twice,and the zero-length segments only once. As there are infinitely fewer zero-length segments than non-zero-length segments, and I am calculating an average, this is a problem that I can safely ignore.
I had originally envisioned (i,j) as one point, and (k,l) as the other point, and did not see how to approach the problem. Fortunately, I used the form above for the numerical solution, which seems to indicate a simpler integral.
Let us address the simpler problem of "What is the average length of a segment within the unit line segment?" and keep track of the statistics. This problem is more easily understood by considering individual points for the "outer integral" and what the lengths are for the "inner integral".
When the outer integral as near the center, the inner integral yield lengths equally likely between 0 and 0.5 (one each way). When the outer integral variable is at an end, the inner integral value yields lengths equally likely between 0 and 1.
When the outer integral variable has a value x<0.5, the inner integral yeilds 2 solutions for numbers less than x, and one for answers between x and 1-x.
These indicate a linear ramp of "x-lengths" from a maximum likelihood of 1.0 at lenth zero, to a likelihood of 0.0 at length one. Multiplying this by the line length yields the integral of 2(1-x)(x), or 2(x-x²). The integral of 2x over 0..1 is one. The integral of 2x² over 0..1 is 2/3. The average length of a segment on a unit segment is 1/3.

The simple linear ramp found in the integral above implies a fairly simple two-variable integral for the average length of a segment in a square:
4ʃʃ(1-x)(1-y)√(x²+y²)dxdy, both integrals being over 0..1.
After looking for a couple of hours, I’ll admit that I do not see an obvious closed-form solution. Unless I find one in the thread, I’ll probably search for one on and off for the next two or three months.