Solution: In order to show that the subspace topology is the discrete topology, we need to show that every singletons are open in the subspace topology. Note that \begin{align*} \{ 1 \} & = A \cap \left( \frac{3}{4}, 2 \right). \end{align*} For any $n \geq 2$, define \[ a_n = \frac{\frac{1}{n} + \frac{1}{n+1}}{2} \text{ and } b_n = \frac{\frac{1}{n} + \frac{1}{n-1}}{2}. \] Then we have \[ \left\{ \frac{1}{n} \right\} = A \cap \left( a_n, b_n \right). \] Thus, each singleton set is open and hence the subspace topology is the discrete topology.