Although it says here that, in the topology on the unit square representing the Möbius loop, the boundary consists of those points of the form $(a,a)$, I cannot find any justification as to why this must be true. Though I have attempted to prove this by proving that it is the unbounded Möbius loop that is formed by the same topology in the absence of the points of this form, this line of attack has not gotten me very far.
So, how may this be proven?
The section pertaining to which the question has been asked is quoted below:
A torus can be constructed as the square [0,1]×[0,1] with the edges identified as (0,y)∼(1,y) (glue left to right) and (x,0)∼(x,1) (glue bottom to top). If one then also identified (x, y) ~ (y, x), then one obtains the Möbius strip. The diagonal of the square (the points (x, x) where both coordinates agree) becomes the boundary of the Möbius strip