Solution: Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that \[ d(x_1, x_2) < \delta \implies \left\vert d(x_1, Y) - d(x_2, Y) \right\vert < \epsilon . \] Given $x_1 \in X$, the number $d(x_1, X) + \epsilon /2$ is not a lower bound of the set $\left\{ d(x,y): y \in Y \right\} $. Thus, there exists $y_1 \in Y$ such that \begin{equation}\label{eq:15Aug2024-1} d(x_1, y_1) < d(x_1, Y) + \frac{\epsilon}{2}. \end{equation} Therefore, given any $x_2 \in X$, with $d(x_1, x_2) < \frac{\epsilon}{2} $, we have \begin{align*} d\left( x_2, Y \right) & \leq d(x_2, x_1) + d\left( x_1, y_1 \right) \\ & < \frac{\epsilon}{2} + d(x_1, Y) + \frac{\epsilon}{2} \\ & < \epsilon + d(x_1, Y). \end{align*} Thus, $d(x_2, Y) - d(x_1, Y) < \epsilon $. Similarly, by changing the role of $x_1$ and $x_2$, we can obtain \[ d(x_1, Y) - d(x_2, Y) < \epsilon. \] Therefore, we proved that \[ d(x_1, x_2) < \frac{\epsilon}{2} \implies \left\vert d(x_1, Y) - d(x_2, Y) \right\vert < \epsilon . \]