First, as $d(x,x)=0$ for all $x\in X$ , it follows that is a cover. Second, suppose and $z\in B_1\cap B_2$ . We claim that there exists a such that \begin{eqnarray} \label{ii} z&\in& B_3\subseteq B_1\cap B_2. \end{eqnarray}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 $$ d(x_1, z)<\varepsilon_1, \quad d(x_2, z)<\varepsilon_2. $$ Now we can define $\delta = \min\{ \varepsilon_1-d(x_1, z), \varepsilon_2-d(x_2, z)\}>0$ , and put $$ B_3 = B_\delta(z). $$ If $y\in B_3$ , then for $k=1,2$ , we have by the triangle inequality \begin{eqnarray*} d(x_k,y) &\le & d(x_k, z) + d(z,y) \\ &< & d(x_k, z) + \delta \\ &\le & \varepsilon_k, \end{eqnarray*}so $B_3\subseteq B_k$ and condition holds.