We have trivially established that one of the sides needs to be, at minimum, cos(π/4). The question is whether the other side can be of length less than 1.

A rectangle with diagonal 1 will have a height cos(θ) and a width sin(θ) for some θ. I'm not going to explain why, but the critical rectangles for us are the ones with an angle between 0 and π/8. You can check up to π/4 if you want, but they fit trivially.

So there are two ways a rectangle can fit inside the box: sitting flush against the edge and diagonally. So if you imagine that the BOX we're trying to fit our rectangle in is situated with vertices at (0,0), (0,H), (sin(π/4),0), and (sin(π/4),H), we can calculate, for each possible rectangle, what the needed height H will be. It will be the minimum of cos(θ) and Q, the highest point of the rectangle when it is sitting skew. So now we have an ugly trig problem and a minimax problem to solve. By the way, you solve the minimax problem by equating Q and cos(θ). Easy peasy.

Wait, so what's this θ, by the way? Situate the rectangle so the long side is along the y-axis, draw the diagonal, it's the smaller internal angle. Okay, great.

Now, tilt your rectangle over so it fits in your box askew. Now the angle between the diagonal of the rectangle and the y-axis is π/4, so the angle the side of your rectangle makes with the y-axis = β = π/4 - θ (why: you have a diagonal of 1 and a base of sin(π/4)). Now draw a bunch of triangles, labeling the right angles, the θ angles, the β angles, and do a lot of projections with sines and cosines to get something like the required Q = sin(θ)sin(β)+cos(θ)cos(β) or something like that, I don't have my scrap paper with me. Which solves for θ = π/12.

I might make some diagrams later or something, but I have some crap I'm supposed to be doing.