Please help me understand how $\overline{\bigcup A_\alpha}\subset \bigcup \overline{A_\alpha}$ does not hold in general. From Munkres Topology section $17$ on Closed Sets and Limit Points.

I have seen answers that say if we pick some neighborhood of $x$ it may intersect some $A_\alpha$, but another neighborhood of $x$ may not intersect this same $A_\alpha$.
However, why do we care if two different neighborhoods do not intersect the same $A_\alpha$ as long as it intersects some $A_\alpha$? After all, we are trying to show that $x$ is in the union of the closures.
 A: Every open neighborhood of $x$ intersects the union; therefore intersects some $A_\alpha$. But a smaller open neighborhood may intersect $A_\beta$ and a smaller one than that may intersect $A_\gamma$, etc. and $\alpha,\beta,\gamma,\ldots$ can be distinct.  Thus you haven't shown that there's just one $A_\alpha$ that will continue to be intersected by every open neighborhood of $x$ no matter how small the neighborhood gets.  In other words, you haven't shown that there is some $\alpha$ for which $x$ is in the closure of $A_\alpha$.
Identifying this flaw in the proof doesn't mean the proposition is false; it only means you haven't proved it's true.  To show it's false you'd use a counterexample.  Here is a counterexample:
$$
A_n = \left( \frac 1 n, \frac 1 n + \frac 1 {n^2} \right).
$$
Then $\displaystyle 0 \in \overline{\bigcup_{n=1}^\infty A_n}$ but $\displaystyle 0 \notin \bigcup_{n=1}^\infty \overline{A_n}.$
The counterexample may shed some light on the fallacy of the argument.  For every $\varepsilon>0$, you have some $n$ so large that the neighborhood $(-\varepsilon,+\varepsilon)$ of $0$ intersects the union.  But there is no $A_n$ that intersects the neighborhood $(-\varepsilon,\varepsilon) = \left( -\frac 1{2n}, +\frac 1 {2n} \right).$  That neighborhood separates $0$ from $A_n$, so that $0$ is not in the closure of $A_n$.
A: As let be $x$  a limit point of $\bigcup A_a$ with $x$ not necessarily in any $A_a$. The neighborhood $U$ of $x$ contains a point $y \ne x$ in $\bigcup A_a$.  But that is only one neighbor hood.  To be a limit point for that one particular $A_a$ every neighborhood of $x$ must intersect that $A_a$.  But you only know that every neighborhood intersects some, maybe different $A_a$.  If each neighborhood intersect different $A_a$ then $x$ is not necessarily a limit point to any.
Consider for example $A_i = \{i/z\mid z \in \mathbb Z-\setminus\{0\}\}$.  Then $\overline{A_i} = A_i$.  And $\bigcup A_i = \mathbb Q$ and $\overline {\bigcup A_i} = \mathbb R \not \subset \bigcup {\overline A_i} = \mathbb Q$.
Consider the "proof" for $x$, $x$ irrational, then every neighborhood $U$ intersects some $A_i$.  Say $x = \pi$ and $3.2 \in N(1/10,\pi)=U$. So $U$ intersects $A_{32}$ as $3.2 = 32/10$ (and $|\pi - 3.2| < 1/10$).  But $N(1/20,\pi)$ need not intersect $A_{32}$.  Indeed  $32/10 - \pi= 3.2 - \pi > .05$ and $\pi - 32/9 = \pi = 2.909090\ldots > .05$.  So $\pi \not \in \overline {A_{32}}$. (But $N(1/20,\pi)$ intersects $A_{31}$... but $N(1/100, \pi)$ doesn't... and so on...)
Indeed every neighborhood does intersect an $A_i$ but there is no $A_i$ that is intersected by all neighborhoods and $\pi$ is not in any $\overline{A_i}$.
