Observation 1 (trivial):
There must be a segment touching each corner of the square
Observation 2 (non-rigorous):
Consider the solution of both main diagonals. Any other solution consisting of exactly 4 segments with a single intersection point has a greater length than the both diagonals.
> This can be seen by moving the intersection point and repeatedly replacing any segment by a polyline of 2 segments. Due to the triangle inequality any of these operations increases the length of the segment set. Note that (well-behaved) curves can be approximated to an arbitrary precision by polylines so this construction is not limited to sets of straight segments (some technicalities are missing for a mathematically rigorous proof).
The solution will be a connected structure. The structure shall thus map to a connected graph of minimal geometrical edge length that links all 4 corners of the unit square.
There is a structure that precisely realizes these needs:
A Steiner tree, which can be seen here:
The total length of segments is approx. 2.732
What is missing for optimality ( or: the dangers of intuition) ?
The proof that the minimal structure must be connected. It seems intuitively obvious that if the structure is disconnected, either a corridor can be found through which rays can pass that intersect two sides of the square or the structure is way too long. This needs to be formalized however to be sure.
After peeking into the other solutions, i found my intuition proved wrong ... The structure need not be connected. However, among the connected ones, the Steiner tree is optimal.