TITLE: Show that $\bigcup \overline{A_\alpha} \subseteq \overline{\bigcup A_\alpha}$
QUESTION [5 upvotes]: Can someone please verify my proof? I am aware that there is a similar question elsewhere, but I need help with my proof in particular.

Show that $$\bigcup \overline{A_\alpha} \subseteq \overline{\bigcup A_\alpha}$$
  Give an example where equality fails.

Let $x \in \bigcup \overline{A_\alpha}$. Then, $x \in \overline{A_\alpha}$ for some index $\alpha$. This implies that every open neighborhood of $x$ intersects $A_\alpha$. But then, every open neighborhood of $x$ intersects $\bigcup A_\alpha$. So, $x \in \overline{\bigcup A_\alpha}$.
To show that equality fails, let $A_k = \left(-1+\frac{1}{2n}, 1-\frac{1}{2n} \right)$. Then, it is clear that$$\bigcup_{k=1}^\infty \overline{A_k} = (-1, 1)$$
and $$\overline{\bigcup_{k=1}^\infty {A_\alpha}} = [-1, 1]$$

REPLY [3 votes]: Another attempt at a counterexample: since the rationals are countable, we can enumerate them by $q_1,q_2\ldots $. Take $A_k=\lbrace q_k\rbrace\subset\mathbb{R}$. Then each $A_k$ is closed since its complement is open. So $\overline{A_k}=A_k$. Hence $$\bigcup\overline{A_k}=\bigcup A_k=\mathbb{Q}$$ 
but $$\overline {\bigcup A_k}=\overline{\mathbb{Q}}=\mathbb{R}$$