Let $x \in X$. If for every net $(x_{\lambda})_{\lambda \in \Lambda}$ in $X$ such that $x_{\lambda} \rightarrow x$ there exists $\lambda_0 \in \Lambda$ such that $x_{\lambda} \in A \ \ \ \forall \lambda \geq \lambda_0$ then $A$ is a neighborhood of $x$.
My solution so far: Suppose that $A \notin \mathcal{N}_x$, where $\mathcal{N}_x$ the family of sets containing $A$, then $x \notin A^o \Leftrightarrow x\in X-A^o=\overline{X-A}$. Since $\overline{X-A}$ Then there exists a net $(x_{\lambda})_{\lambda \in \Lambda}$ in $X$ such that $x_{\lambda} \in \overline{X-A} \ \ \ \ \forall \lambda \in \Lambda$ and $x_{\lambda} \rightarrow x$ which is a contradiction.
I'm not sure if $x_{\lambda} \in cl(X-A)$ indeed is a contradiction.
Another approach is writing down the definition of net convergence and showing that $A \cap U \neq \emptyset$ for every $U \in \mathcal{N}_x$. But this shows (again) that $x\in \overline{A}$, while I would like to show that $x \in A^o$.
Thank you in advance!