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!
If $A$ is not a neighbourhood of $x$, this means that for any open set $O$ that contains $x$, $O \nsubseteq A$. So assume for a contradiction that this is the case. A direct proof is quite doable:
Now let $\Lambda$ be the set of all open sets containing $x$ in reverse inclusion ordering, which makes it into a directed set.
Now for each $O \in \Lambda$ pick $x_O \in O\setminus A$, by the non-inclusion we assumed.
If $O$ is open containing $x$, $O \in \lambda$, and for all $\lambda=O' \ge O$, $x_\lambda \in O' \subseteq O$, so for all $\lambda \ge O $ we have $x_\lambda \in O$. So $x_\lambda \to x$. But by construction, $x_\lambda \notin A$ for all $\lambda$. This contradicts the given property of $A$. So $A$ must be a neighbourhood of $x$ after all.
In your proof you can pick all $x_\lambda \in X-A$ straight away, because of the fact that $x \in \overline{B}$ iff there is a net from $B$ that converges to $x$, as a general fact (part of which I reproved above, in fact).