Suppose we have a space $X$ and a metric $d$ on $X$ . We'd like to show that the metric topology that $d$ gives $X$ is Hausdorff.

Say we've got distinct $x,y\in X$ . Since $d$ is a metric, $d(x,y)\neq 0$ . Then the open balls $B_x = B(x,\frac{d(x,y)}{2})$ and $B_y = B(y, \frac{d(x,y)}{2})$ are open sets in the metric topology which contain $x$ and $y$ respectively. If we could show $B_x$ and $B_y$ are disjoint, we'd have shown that $X$ is Hausdorff.

We'd like to show that an arbitrary point $z$ can't be in both $B_x$ and $B_y$ . Suppose there is a $z$ in both, and we'll derive a contradiction. Since $z$ is in these open balls, $d(z,x) < \frac{d(x,y)}{2}$ and $d(z,y) < \frac{d(x,y)}{2}$ . But then $d(z,x) + d(z,y) < d(x,y)$ , contradicting the triangle inequality.

So $B_x$ and $B_y$ are disjoint, and $X$ is Hausdorff.$\square$