Let $(X,\tau)$ be any generalized topological space. If $A\subseteq X$ is open and $A\neq \emptyset$, it cannot be be strongly nowhere dense: Let $U=A$, then for every nonempty open set $V\subseteq U$ we have $V \cap A = V\cap U = V \neq \emptyset$!

So $A = \emptyset$ is the only strongly nowhere dense subset of $(X,\tau)$.

Let $(X,\tau)$ be any generalized topological space. If $A\subseteq X$ is open and $A\neq \emptyset$, it cannot be be strongly nowhere dense: Let $U=A$, then for every nonempty open set $V\subseteq U$ we have $V \cap A = V\cap U = V \neq \emptyset$.

So $A = \emptyset$ is the only strongly nowhere dense subset of $(X,\tau)$.

Let $X = \omega = \{0,1,2,\ldots\}$ and let $$\tau = \big\{\{1,2\}, \{2, 3\}\big\} \cup \big\{S\subseteq \mathbb{N}: \{1,2,3\}\subseteq S\big\}.$$

Then $(X,\tau)$ is a generalized topological space (but not a topological space, since $\{2\}=\{1,2\}\cap\{2,3\}$ is not a member of $\tau$); moreover $\{0\}$ is strongly nowhere dense.

Let $(X,\tau)$ be any generalized topological space. If $A\subseteq X$ is open and $A\neq \emptyset$, it cannot be be strongly nowhere dense: Let $U=A$, then for every nonempty open set $V\subseteq U$ we have $V \cap A = V\cap U = V \neq \emptyset$!

So $A = \emptyset$ is the only strongly nowhere dense subset of $(X,\tau)$.