Solution: We need to verify the following properties:

1. $d(x,y) \geq 0$ for $x,y \in \mathbb{R} $;
2. $d(x,y) = 0 \iff x = y$;
3. $d(x,y) = d(y,x)$ for $x,y \in \mathbb{R} $;
4. $d(x,z) \leq d(x,y) + d(y,z)$ for any $x,y, z\in \mathbb{R} $.

Note that the first three properties are straight forward. We will discuss the last property (triangle's inequality). Let $x,y,z \in \mathbb{R} $. Then, we need to show that \[ \sqrt{\vert x-z \vert } \leq \sqrt{\vert x - y \vert } + \sqrt{\vert y - z \vert }. \] Note that \begin{align*} \left( \sqrt{\vert x - y \vert } + \sqrt{\vert y - z \vert } \right) ^2 & = \vert x - y \vert + \vert y - z \vert + 2 \sqrt{\vert x - y \vert \cdot \vert y - z \vert } \\[1ex] & \geq \vert x - y \vert + \vert y - z \vert \\[1ex] & \geq \vert x - z \vert \\[1ex] & \geq \sqrt{\vert x-z \vert }. \end{align*} Thus, $d$ is a metric on $\mathbb{R} $.