Following Leonid's suggestion:
(III) $\Rightarrow$ (II): $\bar A$ is closed and $X-(\bar A-A)$ is open, and their intersection is $\bar A\cap(X-(\bar A-A))=\bar A\cap (A\cup(X-\bar A))=A$.
(II) $\Rightarrow$ (I): Let $A=C\cap U$ with $C$ closed and $U$ open. Then $U$ is a neighbourhood of every $x\in A$, and $A\cap U=C\cap U\cap U=C\cap U$; thus $U-(A\cap U)=U-(C\cap U)=U-C=U\cap(X-C)$, which is open, so $A\cap U$ is closed in $U$.
(I) $\Rightarrow$ (III): If $U$ is a neighbourhood of $x\in A$ such that $A\cap U$ is closed in $U$, then $U$ is disjoint from $\bar A-A$; otherwise it would contain a point outside of $A$ each of whose neighbourhoods contains a point of $A$, so $U-(A\cap U)$ couldn't be open in $U$. Thus the union of all the open neighbourhoods contained in all these neighbourhoods for all $x\in A$ is disjoint from $\bar A-A$. It is open and contains $A$. If we add the complement of $\bar A$, which is open, we get the complement of $\bar A-A$, and as the union of two open sets this is open; hence $\bar A-A$ is closed.