I’ll summarize what the open unit balls look like at different points and leave it to you to verify that the descriptions are correct.
It will depend on $|y|$. If $|y|\ge 1$, then of course it is simply $\{x\}\times(y-1,y+1)$. If $y=0$, it’s an open square diamond with vertices $\langle x-1,0\rangle$, $\langle x+1,0\rangle$, $\langle x,1\rangle$, and $\langle x,-1\rangle$. And if $0<|y|<1$, it’s a similar, smaller diamond with a vertical ‘tail’. The centre of the diamond is at $\langle x,0\rangle$, the vertices are at distance $1-|y|$ from the centre, and the tail is the part of $\{x\}\times(y-1,y+1)$ that sticks out of the diamond.