Quotient of the torus being homeomorphic to Mobius strip? 
I'm considering X to be the space $[-1,1] \times [-1,1]$, with left glued to right and top glued to bottom, and points symmetric with respect to the diagonal glued to each other. But I still can't visualize the final shape of X. Is that a cone? And I don't know how this shape could be mapped to a Mobius strip?
 A: "But I still can't visualize the final shape of X. Is that a cone?"
It's actually a cross-cap with a disk removed. We can think of it in terms of surface diagrams. Imagine taking the unit square. Since, as you said, the points symmetric wrt the diagonal are identified, imagine folding the square diagonally in half. You end up with a triangle with one edge the segment between $(-1, 1)$ and $(1, 1)$ and another the edge between $(1, 1)$ and $(1, -1)$. But remember that for a point $(x, 1)$ on the boundary, we have $(1, x) \sim (x, -1)$, so you would have to draw the appropriate arrows to indicate gluing. As a hint, your guess that it is a cone is almost correct; a cone is orientable, but the question implies that $X$ is homeomorphic to a Mobius strip, so $X$ can't be orientable! Think about which way the arrows would point in the surface diagram of a cone and switch the direction of one of them. 
"I don't know how this shape could be mapped to a Mobius strip?"
Here's a cool video showing the homeomorphism in detail: https://www.youtube.com/watch?v=yeaw3BnCs34
You can easily prove this using cut-and-paste topology on the triangular surface diagram constructed earlier. Cut the triangle in half, and reglue to obtain the more familiar rectangular surface diagram for the Mobius Strip.  
