We shall use the this result to prove that $\mathscr{B}$ is a base.

First, as $d(x,x)=0$ for all $x\in X$, it follows that $\mathscr{B}$ is a cover. Second, suppose $B_{1},B_{2}\in\mathscr{B}$ and $z\in B_{1}\cap B_{2}$. We claim that there exists a $B_{3}\in\mathscr{B}$ such that

By definition, $B_{1}=B_{{\varepsilon_{1}}}(x_{1})$ and $B_{2}=B_{{\varepsilon_{2}}}(x_{2})$ for some $x_{1},x_{2}\in X$ and $\varepsilon_{1},\varepsilon_{2}>0$. Then

Now we can define $\delta=\min\{\varepsilon_{1}-d(x_{1},z),\varepsilon_{2}-d(x_{2},z)\}>0$, and put

If $y\in B_{3}$, then for $k=1,2$, we have by the triangle inequality

so $B_{3}\subseteq B_{k}$ and condition 1 holds. ∎