Take some convex polygon $P$. I'm mostly asking about the unit square, but would also appreciate thoughts on general polygons. We want to take a family of line segments inside $P$ that have the same projections as $P$ (that is, any line intersecting $P$ intersects one of the segments). Let $L$ be the minimum total length of such segments. What is $L$?

I have the following bounds:

Upper bound: $L\le\sqrt{2}+\frac{\sqrt6}{2}\approx 2.639$

Take the following construction such that the three segments meet at angles of $2\pi/3$ (the gray dahsed diagonal is not one of the segments in the construction). The sum of the lengths is $\sqrt{2}+\frac{\sqrt6}{2}$.

Lower bound: $L\ge 2$.

The sum of the lengths of orthogonal projections onto the two diagonals is $2\sqrt{2}$. For any segment of length $a$ and angle $\theta$ to the diagonals, the sum of its projections is $a|\sin\theta|+a|\cos\theta|\le a\sqrt{2}$. Thus $L\sqrt{2}\ge 2\sqrt{2}$, hence the lower bound.

I've tried improving the lower bound by using other combinations of projections, or splitting the square into subshapes and using different projections on different subshapes, but couldn't get anything better.