Your answer is correct.
Though it wasn't asked for in your problem, a justification of your answer would be edifying:
The open unit ball is by definition $ B(0,1)=\bigl\{\, (x,y)\in\Bbb R^2\,\bigl|\,|x|+|y|<1\,\bigr\}. $ Towards determining geometrically what $B(0,1)$ is, let's consider cases:
Let's find the portion of $B(0,1)$ in the first quadrant of the Euclidian plane. If $(x,y)$ is in the first quadrant of the Euclidian plane, then $(x,y)\in B(0,1)$ if and only if $x+y<1$. If you sketch the graph of the equation $x+y=1$, you should be able to convince yourself that $Q_1\cap B(0,1)$ is the region depicted below (you want the region in Quadrant 1 with $x+y<1$, this would be the region in Quadrant 1 beneath the graph of the equation $x+y=1$).

If you consider the other cases according to the other three quadrants, you can establish that $B(0,1)$ is as you described and as depicted below (or, you could use a symmetry argument):

Note that since $B(0,1)$ is the open unit ball, points on the the boundary of the diamond above are not in $B(0,1)$.