We have,
$|x|+|y|=1=\begin{cases}x +y=1, & x, y \in \text { I quadrant } \\ -x+ y=1, & x, y \in \text { II quadrant } \\ -x -y=1, & x, y \in \text { III quadrant } \\ x-y=1, & x, y \in \text { IV quadrant }\end{cases}$
Clearly, $A B C D$ is a square.