 Solution

For $(i)\Rightarrow (ii)$, suppose $B=U\cap F$ with $U$ open and $F$ closed. Since $B\subseteq F$, also $\overline{B}\subseteq F$. Therefore $U\cap \overline{B}=B$, showing that $B$ is relatively open in its closure.

For $(ii)\Rightarrow (iii)$, write $B=U\cap \overline{B}$ with $U$ open in $X$. Then $\overline{B}\setminus B=\overline{B}\cap (X\setminus U)$ is an intersection of closed sets, hence closed in $X$.

For $(iii)\Rightarrow (i)$, set $F=\overline{B}$ and $U=X\setminus (\overline{B}\setminus B)$. The assumed condition makes $U$ open; $F$ is closed and $U\cap F=B$. Thus the three conditions are equivalent. They characterize a locally closed subset.