Standard Discrete Metric induces Discrete Topology

Theorem
Let $M = \left({A, d}\right)$ be the discrete metric space on $A$.

Then $d$ induces the discrete topology on $A$.

Thus the discrete topology is metrizable.

Proof
Let $a \in A$.

From Subset of Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$.

So, in particular, $\left\{{a}\right\}$ is open in $\left({A, d}\right)$.

This holds for all $a \in A$.

From Metric Induces Topology it follows that $\left\{{a}\right\}$ is an open set in $\left({A, \tau_{A, d}}\right)$.

The result follows from Basis for Discrete Topology.