Solution: let $X$ contains two distinct points $x$ and $y$. Then by the definition of metric, $r = d(x,y) > 0$. Consider the open balls \[ B_x \coloneqq B\left( x, \frac{r}{4} \right) ,\quad B_y \coloneqq B\left( y, \frac{r}{4} \right) \] around $x$ and $y$, respectively. We claim that the closure of these two balls are disjoint.

Suppose, to the contrary that $z \in \overline{B_x} \cap \overline{B_y} $. This means, for every open neighborhood $U_z$ of $z$, $U_z$ must intersects with $B_x$ and $B_y$. So consider the open neighborhood $U_z = B\left( z, \frac{r}{4} \right) $.

By assumption $U_z$ has a point of $B_x$, say $x' \in U_z \cap B_x$. Look at the figure below. Then we have \begin{align*} d(x, z) \leq d(x, x') + d(x', z) < \frac{r}{4} + \frac{r}{4} = \frac{r}{2}. \end{align*} Since $U_z$ also contain a point of $B_y$, say $y'$. Then by the similar argument we have $d(y,z) < \frac{r}{2}$. But then, \begin{align*} d(x,y) \leq d(x,z) + d(z,y) = d(x,z) + d(y,z) < \frac{r}{2} + \frac{r}{2} = r, \end{align*} a contradiction. Thus, there does not exist any element in the intersection of $\overline{B_x} $ and $\overline{B_y} $ and hence we conclude that $\overline{B_x} \cap \overline{B_y} = \emptyset $.