[proofplan] Given distinct points $x, y$ in a [metric space](/page/Metric%20Space), their distance $r = d(x,y) > 0$ provides a natural scale. Open balls of radius $r/2$ centred at $x$ and $y$ are the required disjoint open neighbourhoods, with disjointness following from the triangle inequality. [/proofplan] [step:Construct disjoint open balls using half the separation distance] Let $x, y \in X$ with $x \neq y$. By the positivity axiom of the metric, $r := d(x, y) > 0$. Define $U = B(x, r/2)$ and $V = B(y, r/2)$. Both are open in the metric [topology](/page/Topology), and $x \in U$, $y \in V$. [/step] [step:Verify disjointness of $U$ and $V$ via the triangle inequality] Suppose for contradiction that $z \in U \cap V$. Then $d(z, x) < r/2$ and $d(z, y) < r/2$. The triangle inequality gives \begin{align*} r = d(x, y) \leq d(x, z) + d(z, y) < \frac{r}{2} + \frac{r}{2} = r. \end{align*} This yields $r < r$, a contradiction. Therefore $U \cap V = \varnothing$, and $(X, \tau)$ is Hausdorff. [guided] The proof is entirely determined by the choice of radius. Why $r/2$? Any radius $\rho$ with $0 < \rho \leq r/2$ would work: we need $\rho + \rho \leq r$ to ensure the triangle inequality gives a contradiction. The choice $\rho = r/2$ is the largest such radius, giving the "tightest" pair of disjoint balls. Suppose $z \in B(x, r/2) \cap B(y, r/2)$. Then $d(z, x) < r/2$ and $d(z, y) < r/2$. The triangle inequality bounds: \begin{align*} d(x, y) \leq d(x, z) + d(z, y) < \frac{r}{2} + \frac{r}{2} = r = d(x, y). \end{align*} We have derived $d(x,y) < d(x,y)$, which is absurd. So no such $z$ exists, and the balls are disjoint. Since open balls are open in the metric topology, $U$ and $V$ are the required separating neighbourhoods. [/guided] [/step]