 Solution

We prove (ii)$\Rightarrow$(iii). If $B$ is open in $\overline{B}$, its complement $\overline{B}\setminus B$ is closed in the subspace $\overline{B}$. A relatively closed subset of a closed subset of $X$ is closed in $X$: it is the intersection of that closed subset with another closed subset of $X$. Since a closure is closed, $\overline{B}\setminus B$ is closed in $X$.