Antipodal gluing of a disk is homeomorphic to a disk. Given $D=\{(x,y)\in \mathbb{R}^2 \mid x^2+y^2\leq 1\}$, if I define $\sim $ as $(x,y)\sim (x',y') \iff x'=-x, y'=-y$ is the gluing of points across the line $y=-x$. The same way that the antipodal gluing of a circle is homemorphic to a circle, the same is true for a disk. The problem is, I am not really sure what a continuous map from the disk to the glued disk would look like--let alone a homeomorphism. The plan I had was to identify points over the reflective line $y=-x$ but I am having trouble writing down a closed form. Any hints?

If you identify points $(x,y)$ and $(-x,-y)$ then you are not "glueing points across the line $y=-x$" but extending the antipodal map to the interior of $D$. In terms of (vectors or) complex numbers $z$ you are identifying $z$ and $-z$. Let $\hat z:=\{z,-z\}$ be the equivalence class represented by $z$, and denote the resulting space by $\hat D$. A homeomorphism $f:\>\hat D\to D$ is then given by
$$f\bigl(\hat z):=z^2\qquad\bigl(\hat z\in\hat D)\ .$$
Think about it!
