(related to Proposition: Metric Spaces are Hausdorff Spaces)
Set \(\epsilon:=\frac 12 d(x,y)\). Because by assumption \(x\neq y\) , we have \(d(x,y)>0\). The open balls
\[U:=B(x,\epsilon),~~~~~~~~~~~~V:=B(y,\epsilon)\] are by definition neighborhoods of \(x\) and \(y\). We shall show that \(U\cap V= \emptyset\).
Assume that \(U\cap V\neq \emptyset\) and let \(z\in U\cap V\) be a point in both neighbourhoods. Then it would follow that
\[2\epsilon=d(x,y)\le d(x,z)+d(z,y) < \epsilon+\epsilon,\]
which is a contradiction. Therefore we have indeed \(U\cap V= \emptyset\), which means that the metric space \((X,d)\) is a Hausdorff space.
