Informally, the proof is to take a square and let 2 opposite corner A and C run around the loop. Doing that, you rotate the square while trying to fit the 2 other corners B and D within the loop. If you start with A and C on a horizontal line, and B and D can fit within the loop, then after 1/4 turn, A and C are on a vertical line, where BD originally was and now B and D are outside of the loop.
So, at some point (for some angle between AC and the horizontal), you go from "B and D both fit in the loop" to "B and D fit outside the loop". At that point, it should be possible to fit both B and D on the loop.
There are plenty of loose ends, but it makes me feel such a square must exist.