If \((X,d)\) is a metric space, then it is Hausdorff [see number 8].
Solution
Let \(x\ne y\) in the metric space \((X,d)\), and put \(r=d(x,y)/3>0\). If a point \(z\) belonged to both \(B(x,r)\) and \(B(y,r)\), then the triangle inequality would give \[ d(x,y)\le d(x,z)+d(z,y)<2r=\frac23d(x,y), \] a contradiction. Thus \(B(x,r)\) and \(B(y,r)\) are disjoint open neighborhoods of \(x\) and \(y\). Hence every metric space is Hausdorff.