Consider two diagonally opposite points, A and C, and draw a circle with those points at opposite ends of its diameter. One of my earlier posts pointed out that A and C must lie on opposite (parallel) lines of the square, so the solution must have two parallel lines of the square going through A and C. Consider the following figure - if those two parallel lines are not at 90 degrees to line AC, then they will intersect the circle at two points, once at point A or C and once at some other points which I'll call X and Y. Angle A-Y-C is a right angle, which I've proven in a previous post, meaning that line AY is the length of the edges of the square.

In the next figure, the red circle is essentially the same as the black circle in the figure above, except that it's been moved to the right. Specifically, the red circle is drawn such that its perimeter includes the intersection of lines AC and BD, and its center is on the (infinite) line AC. Based on the previous paragraph, if you were to draw a line from the intersection of AC BD to another point on the red circle, then its length would be the edge length of the square if you were to draw lines of the square at that angle.

If you do the same thing with points B and D, that will produce the blue circle. Now what we need to find are two lines, one going from the AC BD intersection to the red circle and one going from the AC BD intersection to the blue circle, which are perpendicular and equal in length so they would produce a square. In order to pull that off with a compass and straight edge, you can rotate the blue circle 90 degrees around the AC BD intersection, which creates the orange circle in the figure. Then the red circle and orange circle will intersect at one point* labeled E, and the line from the AC BD intersection to point E will be the length of the square's edges and will be parallel to the square's lines going through points B and D. It would then be straightforward to use that line to construct the square.

I did not walk through how to actually do all that with a compass and straight edge, but it can all be done.

*Footnote: This is the exception I alluded to before the spoiler. If the red circle and the orange circle have no intersection (aside from the AC BD intersection itself), then there will be no solution. That will be the case if and only if AC and BD are perpendicular and of different lengths.