Now, I have the impression that this section is rather muddled. Having tried to find a solution myself, I would like to display a coherent, short solution which is - in essence - a detailed version of P.SUNDARAM's beautiful answer and then bring this HINT by CAMERON to a close by displaying how to use the fact he stated.
Solution not using Cameron's hint:
Assuming a non-empty, open subset $U$ of $\overline{A \cup B} = \bar{A} \cup \bar {B}$, let $V = U\cap (X-\bar{A})$. Then $V$ is open and non-empty - the latter because otherwise $U\subset\bar{A}$, contradicting the assumption that $A$ is nowhere dense. Then $V\subset\bar{B}$, but that is impossible by the same reasoning. So we reached a contradiction to the assumption that $U$ was non-empty, meaning that any open subset of $\overline{A \cup B}$ is empty.
Using Camerons hint:
Assuming a non-empty, open subset $U$ of $\overline{A \cup B}$, there would exist non-empty, open subsets $V_1$ and $V_2$ of $U$, s.t. $V_1\cap A = \emptyset = V_2\cap B$, by the hint. But any element in $\overline{A}$ has the property that any open neighborhood if it has non-empty intersection with $A$. But $V_1$ is an open neighborhood of any of its points, so $V_1\subset \overline{B}$ and $V_2\subset \overline{A}$. But this contradicts the assumption that $A$ and $B$ are nowhere dense. So $U$ must be empty, and hence the interior of $\overline{A \cup B}$ must be empty as well.