HINT: Suppose that $A$ is nowhere dense. You want to show that if $U$ is a non-empty open set in $X$, then there is a non-empty open set $V\subseteq U\setminus A$. You know that $\operatorname{cl}A$ does not contain any non-empty open set, so subtracting it from an open set shouldn’t remove very much. You know that if $W$ is open and $C$ is closed, then $W\setminus C$ is open, so $U\setminus\operatorname{cl}A$ is open. Does it work for $V$?
Now suppose that every non-empty open subset $U$ of $X$ has a non-empty open subset disjoint from $A$; you want to show that $\operatorname{cl}A$ has empty interior. That is, you want to show that $\operatorname{cl}A$ does not contain any non-empty open set. Suppose, on the contrary, that $\varnothing\ne U\subseteq\operatorname{cl}A$, where $U$ is open. Apply your hypothesis to $U$ to get an immediate contradiction.