Pseudometric Space is Metric Space iff Kolmogorov

Theorem
Let $M = \left({S,d}\right)$ be a Pseudometric Space.

Let $T=\left({S,\tau}\right)$ be the topological space over $S$ induced by $d$.

Then $M$ is a Metric Space iff $T$ is a Kolmogorov Space.

Proof
If $M$ is a metric space, then since Metric Space is Hausdorff, $M$ is trivially Kolmogorov.

Suppose now that $M$ is a Kolmogorov space.

Let $a,b \in S$ such that $a \ne b$.

Then there must be a $U \in \tau$ that distinguishes $a$ from $b$. That is, such that
 * $((a \in U) \land (b \notin U)) \lor ((b \in U) \land (a \notin U))$.

Suppose without loss of generality that $a \in U$ and $b \notin U$.

Then by the definition of the induced topology, there must be an $\epsilon \in \R_{>0}$ such that the open ball $B_\epsilon (a) \subset U$.

Since $b \notin U$, we must have $d(a,b) \ge \epsilon$.

Since $\epsilon > 0$ we see that $d(a,b)>0$.

Since this holds for any pair of distinct points, $M$ is a metric space.