The problem of the proof is that for different $U$ ’s the set $A_{\alpha}$ may be different, so that we actually did not show that every neighborhood $U$ of $x$ intersects some particular $A_{\alpha}$ , we showed that every neighborhood $U$ of $x$ intersects some $A_{\alpha}$ where $\alpha$ may depend on the choice of $U$ , and we cannot conclude from this that $x$ lies in the closure of some particular $A_{\alpha}$ . As an example, we use the one from the previous exercise, $A_{n}=\{\tfrac{1}{n}\}$ , $A=\cup_{n\in\mathbb{Z}_{+}}A_{n}$ . $0\in\overline{A}$ , and $I_{n}=(-\tfrac{1}{n},\tfrac{1}{n})$ contains $0$ and intersects $A$ at points $\tfrac{1}{m}$ for $m>n$ , so $I_{n}$ does, indeed, intersect sets $A_{m}$ for $m>n$ , but no set $A_{m}$ intersects every $I_{n}$ .