I assume from what you're saying you are looking at the square from the outside?

1 & 2 are definitely true, but how do you prove it mathematically....

I'd start with the fact that the square has rotational symmetry, so your viewpoint has two possible extremes: ie with a side facing you, or with a corner facing you. Any other position falls between these two. If a side faces you, perspective dictates that the opposite side will appear shorter and therefore be fully obscured, and the two other sides will be receding behind the "front" side, and also fully obscured. Therefore in this position you can see only one side of the square. In fact this face doesn't have to be exactly perpendicular to see only one face - if it is tilted either way slightly, so that the "side" is aligned such that the front and rear corner of it are in line from your perspective, it would still be invisible - ie there is a small range of angles from which only one side is visible.

With a corner facing you, by definition the corner closest to you is the point at which the two visible sides converge, and the opposite corner is the point where the invisible sides converge. You can only see the two sides on the same principle of perspective as in the first instance.

Now you can rotate the square through 45 degrees, at which one side becomes progressively foreshortened until the side is (nearly) perpendicular to you. The foreshortened side has now been obscured, and the opposite side has not emerged, which is the position of seeing only one side. Due to the rotational symmetry, there is no other viewpoint that is not represented in this 45 degree swing, therefore no other will exist at which any more than two sides will be visible. If you need to express it numerically, subtend an arc that defines the limits of the subjective "width" of the square, the further side always has a smaller arc than the nearer side (ie perspective in numbers....) when face-on.

When rotating the square, the point of the corner is always closest to you except in the one few circumstance(s) where the plane is (nearly) perpendicular to your line-of-sight. The "sides" of the square from your viewpoint will disappear from view at a range of angles corresponding to the arc of your angle of vision, so when you can see only one face, that face is positioned tangential to the radius from your position to the face. As soon as the corner becomes closer than the other parts of the face (ie the corner breaks into that radius), then a second side will become visible, confirming that no 2 is true.

Not sure what exactly you mean in 3 - it looks like a statement of the obvious, but if you want some other way to express it, I guess you could easily prove that the closer object intersects all radii from you to the further object.

Hope you're less confused than I am now