1Before going to the stated problem... I would rather prove a property of linear equations, which I would use in the required proof.
Proposition: Every system of linear equations has either no solution, one solution or infinite solutions (i.e. 2 distinct solutions mean infinite solutions)
Proof: Let A be a mxn matrix, b ε Rm and u,v ε Rn, u≠v such that Au=b=Av (Solution of AX=b.... system of linear equations).
Au=b=Av implies A(u-v)=0 implies A(α(u-v))=0 for all α ε R
This means that u+α(u-v) is also a solution.
Hence Proved.
NOW, the problem at hand can be reframed in terms of linear equations.
Let the co-ordinates of the vertices of the inscribed squares be (x1,x2),(x3,x4),(x5,x6) and (x7,x8)
and let the equations of the sides of the quadrilateral on which they lie be a1x+b1y=c1,a2x+b2y=c2,a3x+b3y=c3 and a4x+b4y=c4.
The vertices satisfy this and we get the system of linear equations :
a1x1+b1x2+0(x3+ x4 + x5 + x6 + x7 + x8) = c1 ,
a2x3+b2x4+0(x1+ x2 + x5 + x6 + x7 + x8) = c2 ,
a3x5+b3x6+0(x3+ x4 + x1 + x2 + x7 + x8) = c3 and
a4x7+b4x8+0(x3+ x4 + x5 + x6 + x1 + x2) = c4
Now if two distinct squares can be inscribed it implies that the system of equations has two distinct solutions which by the previous theorem implies that there are infinite solutions, i.e. infinite squares of different lengths can be inscribed in the quadrilateral.