$x \in \bar A$ $\iff$ $\exists (x_\lambda)_{\lambda \in \Lambda} \subset A$ net with $(x_\lambda) \to x$
What I did:
$\Leftarrow$: Let $(x_\lambda) \subset A$ a net with $(x_\lambda) \to x$. Let $V$ an open set with $x \in V$, then $\exists \lambda_0 \in \Lambda$ with $\forall \lambda \geq \lambda_0$, $x_\lambda \in V$. But $x_{\lambda} \in A, \forall \lambda \geq \lambda_0$. Then $A \cap V$ isn't empty. So $x \in \bar A$
$\Rightarrow$: I don't know how to proceed in this side. Could someone give a hint of what net I should get?
Take as the indexing family $\Lambda$ the family of all open neighbourhoods of $x$, ordered by inclusion. Since $x\in \bar{A}$, there is for every $U\in\Lambda$ an $x_U\in A\cap U$. The net $x_U$ converges to $x$.