Prove that $\overline{A\cup B} = \overline{A} \cup \overline{B} $ where $\overline{A}$ is the closure of $A$ I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I would like to ask, if this constitutes a valid proof of the fact the closure of the union of two sets equals the union of the respective closures?

$$\overline{A \cup B} = \overline{A} \cup \overline{B}$$

Proof.
Suppose $\displaystyle x\in \overline{A\cup B} .$ Let $\displaystyle L$ be the set of limit points $\displaystyle A\cup B$. Then, either $\displaystyle x\in A\cup B$ or $\displaystyle x\in L$. If $\displaystyle x\in A\cup B$, then $\displaystyle x\in A$ or $\displaystyle x\in B$ or $\displaystyle x$ belongs to both. So, $\displaystyle x\in \overline{A} \ \cup \overline{B}$. If $\displaystyle x\in L$, then $\displaystyle x$ is a limit point of atleast one of the sets, $\displaystyle A$ or $\displaystyle B$. ($\star$)
Thus, $\displaystyle x\in \overline{A} \cup \overline{B}$.
Altogether, $\displaystyle \overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$.
In the opposite direction, suppose $\displaystyle x\in \overline{A} \cup \overline{B}$. Then, $\displaystyle x\in \overline{A}$ or $\displaystyle x\in \overline{B}$ or $\displaystyle x$ belongs to both. Since $\displaystyle A\subseteq A\cup B$, $\displaystyle \overline{A} \subseteq \overline{A\cup B}$. Similarly, $\displaystyle \overline{B} \ \subseteq \overline{A\cup B}$. Consequently, $\displaystyle x\in \overline{A\cup B}$.
Edit:
$\star$ If $x$ is a limit point of $A \cup B$, then for all $\epsilon>0$, by definition, the open interval $(x-\epsilon,x+\epsilon)$ intersects $A \cup B$ in some point other than $x$. Consequently, $V_\epsilon(x)$ intersects atleast one of $A$, $B$ in some point other than $x$. Thus, $x$ is limit point of atleast one of $A$, $B$.
 A: You don't have to use limit points at all: $\overline{A}$ and $\overline{B}$ are both closed and their union is closed and contains $A \cup B$. So $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$ is immediate (the closure is the minimal  closed superset).
The reverse follows from $A \subseteq A \cup B$ so $\overline{A} \subseteq \overline{A \cup B}$ and ditto for $B$ and taking these two inclusions together we get $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$, and we have equality.
Your proof under $\star$ is wrong. To have $p \in L=(A \cup B)'$ means $$\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (A \cup B) \cap (p-\varepsilon, p+\varepsilon))$$ while $p \in A' \cup B'$ means
$$(\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (A \cap (p-\varepsilon, p+\varepsilon)))\\
 \lor (\forall \varepsilon>0: \exists q: (q \neq p) \land (q \in (B \cap (p-\varepsilon, p+\varepsilon)))$$ so you first have to decide whether every pointed neighbourhood of $p$ is going to intersect $A$ all the time or $B$ all the time. You make it seem like you can just go from the former to the latter, while this is a logical fallacy.
To illustrate how a valid proof using limit points would go:
let $x \in \overline{A \cup B}$. Then either $x \in A \cup B$ so $x \in A$ or $x \in B$ and in either case it's clear that $x \in \overline{A} \cup \overline{B}$, so assume $x \notin A \cup B$ but $x \in (A \cup B)'$. Suppose that $x \notin \overline{A}$, so in particular $x \notin A$ and $x \notin A'$ so then we can find a $\varepsilon>0$ so that $(x-\varepsilon, x+\varepsilon) \cap A = \emptyset$. But if $\delta>0$ is arbitrary, let $\delta'  =\min(\delta, \varepsilon)>0$ and note (as $x \in A \cup B)'$, the open set $(x-\delta', x+\delta')$ must intersect $A \cup B$ in some $q$. Then as $\delta' \le \varepsilon$, $q \notin A$ by the choice of $\varepsilon$ and so $(x-\delta, x+\delta)\supseteq (X-\delta', x+\delta')$ actually intersects $B$ and as $\delta$ was arbitrary, $x \in B'$ and so $x \in \overline{B}$ as required. So $x \in \overline{A \cup B}$ implies $x \in \overline{A} \cup \overline{B}$. The reverse inclusion is already done OK in the OP's post.
A: There is just a subtle point. If you have $x\in L$, then, by definition, every neighborhood of $x$ intersects $A\cup B$ at a point different from $x$.
But in order to state that $x$ is a limit point of $A$ you need that every neighborhood of $x$ intersects $A$ at a point different from $x$ (similarly for $B$, of course).
Now, if every neighborhood of $x$ intersects both $A$ and $B$ at points different from $x$, you have that $x\in\bar{A}$ and $x\in\bar{B}$, so no problem in stating $x\in\bar{A}\cup\bar{B}$.
Suppose that a neighborhood $U$ of $x$ intersects $A$ at a point different from $x$, but does not intersect $B$ at points different from $x$. We want to show that $x$ is actually a limit point of $A$. Indeed, let $V$ be any neighborhood of $x$. Then also $U\cap V$ is a neighborhood of $x$, and so it intersects $A\cup B$ at a point different from $x$. This point cannot be in $B$, because $(U\cap V)\cap B\subseteq U\cap B\subseteq\{x\}$. So $V$ actually intersects $A$ at a point different from $x$.
Similarly for the remaining case where $U$ intersects $B$ but not $A$.
A: Restated:
Let $x \in L.$
Then every punctured neighbourhood of $x$, $N_r(x),$ contains a point of $A \cup B$, i.e. of $A$ or $B.$
Assume $x \not \in \overline {B}.$
Then there is an $s>0$ sucht that $N_s(x) \cap B =\emptyset. $
This implies that for all $t<s$:
$N_t(x) \cap B=\emptyset.$
Hece $N_t(x)\cap A \not =\emptyset$ which
implies $x \in \overline{A}.$
Similarly, if
$x \not \in \overline {A}$ we get $x \in \overline{B}. $
Thus
$\overline{A\cup B} \subset \overline{A}\cup \overline {B}. $.
