Metric spaces are Hausdorff

Statement

Let $(X, d)$ be a metric space, with the metric topology. Then $X$ is a Hausdorff space.

Remarks

Proof: for $x \neq y$, let $r = d(x, y) / 2 > 0$. The open balls $B(x, r)$ and $B(y, r)$ are open. They are disjoint: a point $z$ in both would give $d(x, y) \le d(x, z) + d(z, y) < 2r$, by the triangle inequality. Thus each metrizable space is Hausdorff. A non-Hausdorff space, such as an infinite set with the cofinite topology, is not metrizable. For a pseudometric, the statement fails: points at distance $0$ cannot be separated.

Explanations

L1 human

Two distinct points of a metric space are at a positive distance. Balls of half that radius about them do not meet.