(Kelley's General Topology) Exercise G chapter 1. I am finding difficultes in solving the following exercise written on the Kelley's book as in the title. Could anyone help me?
Thanks in advance.
If $A$ is dense in a topological space and $U$ is open, then $U \subseteq \overline{(A \cap U)}$.

Suppose that $x\in U$. Since $A$ is dense, $x$ is contained in the closure of $A$. It follows that there exists a net $(x_{\alpha})_{\alpha\in D}$ in $A$ (where $D$ is some index set directed by a relation $\geq$) such that $x_{\alpha}\to x$. By the definition of convergence of nets, since $x\in U$ and $U$ is open, there exists some $\alpha_{0}\in D$ such that $\alpha\geq \alpha_0$ implies that that $x_{\alpha}\in U$. Hence, $(x_{\alpha})_{\alpha\geq\alpha_0}$ is a net in $A\cap U$ that also converges to $x$. It follows that $x$ is in the closure of $A\cap U$. Since $x\in U$ was arbitrary, the conclusion that $U\subseteq\overline{A\cap U}$ follows.

Alternative proof, with no reference to nets $\phantom{---}$Let $x\in U$. Since $x$ is in the closure of $A$, either $x\in A$ or $x$ is an accumulation point of $A$; see Theorem 1.7 in Kelley (1955, p. 42).
Case 1 $\phantom{---}$$x\in A$. Then, $x\in A\cap U\subseteq\overline{A\cap U}$, trivially.
Case 2 $\phantom{---}$$x$ is an accumulation point of $A$. I will show that $x$ is an accumulation point of $A\cap U$. Let $V$ be an open set containing $x$. Then, $U\cap V$ is also an open set containing $x$. Therefore, $A\cap (U\cap V)\setminus\{x\}$ is not empty because $x$ is an accumulation point of $A$. Since $V$ was an arbitrary open neighborhood of $x$, it follows that $x$ is an accumulation point also of $A\cap U$—remember that $(A\cap U)\cap V\setminus\{x\}$ is not empty. Hence, $x$ is in the closure of $A\cap U$.
Conclusion: $U\subseteq\overline{A\cap U}$.
