Show that $\mathbb{R}$ is Hausdorff.
The argument is not complete, until you find $\varepsilon$ and $\delta$ such that $$ (x-\varepsilon,x+\varepsilon)\cap(y-\delta,y+\delta)=\emptyset $$ Since $x\ne y$, it is not restrictive to assume $x<y$. In order the intersection above is empty, it's sufficient that $$ x+\varepsilon<y-\delta $$ that is, $\varepsilon+\delta<y-x$. Take $$ \varepsilon=\delta=\frac{y-x}{3} $$ and you're done.
Assume $x <y$. Then $$x<\frac{x+y}{2}<y$$
Here the sets $(-\infty,\frac{x+y}{2})$ and $(\frac{x+y}{2},\infty)$ makes the separation for $x$ and $y$