The Open Subsets of a Metric Space Form a Topology

In the setting of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness, let $(X,d)$ be a metric space, and let $\mathcal{T}_{d}$ be the set of all $U\in\mathcal{P}(X)$ that are open in $(X,d)$.

(i) $\mathcal{T}_{d}$ is a topology on $X$, so $(X,\mathcal{T}_{d})$ is a topological space.