Solution: We first claim that the set $A$ is not bounded with respect to the usual metric. For that, we will prove that $A$ is not contained any ball $B(x,r) = (x-r, x+r)$ for $x \in \mathbb{R} $ and $r > 0$. Take $y = \vert x \vert + r$. Since $\vert x \vert + r > r$, so $y \in A$ but clearly $y \notin B(x,r)$. Thus, $A$ is not bounded with respect to the Euclidean metric.

On the other hand, with respect to the discrete metric the set $A$ is bounded as $A$ is contained in the ball $B(0,2) = \mathbb{R} $. To see that the ball $B(0,2) = \mathbb{R} $, note that \begin{align*} B(0,2) = \left\{ x \in \mathbb{R} : d(x,0) < 2 \right\}. \end{align*} As $d$ is the discrete metric, distance between any two points is less than or equals to $1$. Thus, $A$ is bounded.