Solution: Let us determine the balls in $A$ with the induced metric. Let $\mathbf{p} = (a,b)$ be the center of the ball. Then, \begin{align*} B_r^A(\mathbf{p} ) & = \left\{ (x,y) \in \mathbb{R} ^2: d(\mathbf{p} , (x,y)) < r \right\} \cap A \\ & = \left\{ (x,y) \in \mathbb{R} ^2: (x-a)^2 + (y-b)^2 < r^2 \right\} \cap A. \end{align*} Take $r= 1$ and $p = (0,0)$, then ball $B_r^A(\mathbf{p} )$ is the union of a quarter ball and a line segment joining $(-1,1)$ and $(1,-1)$.

If we take a point $q$, other than the origin, on the line segment $y = -x$ then we can always choose $r>0$ such that the ball $B_r^A(q) = B_r(q) \cap A$ is a part of the line segment (see figure below). Finally, if we take a point $s$ in first quadrant and the radius is less than the Euclidean distance of the point from the $x$-axis and the $y$-axis, then the ball in $A$ is same as the ball in $\mathbb{R} ^2$. The other types of balls can also be achieved by taking points, for example, $(1,0)$ or $(0,1)$, and take $ r = \frac{1}{2}$, then the ball is half ball.