Closure of union of two sets Show that
$$\overline{A\cup B} = \overline A\cup\overline B$$
$\overline A=A\cup A'$ where $A'$ are the limit points
My attempt: 
Since $\overline A$ is closed and $\overline B$, the union of two closed sets is closed.
Hence $\overline A\cup \overline B$ can be rewritten as $\overline{A\cup B}$
My Second attempt: 
$\overline{A\cup B}=(A\cup B)\cup (A\cup B)'=(A\cup B)\cup (A'\cup B')=A\cup A' \cup B \cup B' = \overline{A}\cup \overline{B}$
 A: You have only proved that $\overline A\cup\overline B$ is closed. You must prove two things more:


*

*$A\cup B\subset\overline A\cup\overline B$. This is very easy.

*Every closed set that contains $A\cup B$ also contains $\overline A\cup\overline B$. This is not so easy.


Some hints for the second part: Consider a closed set $F$ that contains $A\cup B$. Suppose now that there is some $x\in\overline A\cup\overline B$ such that $x\notin F$. Then $x$ can be in $\overline A$ or in $\overline B$.
If it is in $\overline A$, for example, consider an open neighbourhood $U$ of $x$. Now, since $x\notin F$, consider an open neighbourhood $V$ of $x$ such that $F\cap V=\emptyset$. What is $U\cap V$? Is empty $U\cap V\cap A$? Can you arrive at a contradiction?
The other possibility is that $x\in\overline B$. Can you make the same reasoning?
A: Note that closures preserve set inclusion, i.e. if $A \subset B$ then $\overline A \subset \overline B$.
We want to show $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $A \subset A\cup B$ and $B \subset A\cup B$ so $\overline{A} \subset \overline{A\cup B}$ and $\overline{B} \subset \overline{A\cup B}$. So, $\overline{A} \cup \overline{B} \subset \overline{A\cup B}$. On the other hand, $A \subset \overline{A}, B\subset \overline{B}$, so $A\cup B \subset \overline{A} \cup \overline{B}$. The latter is closed, so $\overline{A\cup B} \subset \overline{A} \cup \overline{B}$.
We conclude $\overline{A\cup B} = \overline{A} \cup \overline{B}$. $\ \  \blacksquare$
A: Your second attempt only needs one more fact: $(A \cup B)' = A' \cup B'$.
To see this: pick $x \in A'$. Then every neighbourhood $U$ of $x$ intersects $A$ in at least one point different from $x$. This point is also in $A \cup B$, so then every neighbourhood of $x$ intersects a point of $A \cup B$ not equal to $x$, so $x \in (A\cup B)'$. Similarly, if $x \in B'$, $x \in (A \cup B)'$, so we have $A' \cup B' \subset (A \cup B)'$.
On the other hand, if $x \in (A \cup B)'$, we need to show $x$ is in $A' \cup B'$. So assume not, $x \notin A'$, which means there is a neighbourhood $U$ of $x$ that intersects $A$ in an empty set or just $\{x\}$. Also $x \notin B'$, so there is a neighbourhood $V$ of $x$ such that $V$ intersects $B$ either in the empty set or $\{x\}$. Now, $U \cap V$ is a neighbourhood of $x$, and $(U \cap V) \cap (A \cup B) = (U \cap V \cap A) \cup (U \cap B \cap V) \subset (U \cap A) \cup (V \cap B)$, which is also either the empty set or $\{x\}$, from the properties of $U$ and $V$. But this contradicts that $x \in (A \cup B)'$. So our assumption was incorrect: $x$ must be in $A' \cup B'$, as required.
A: You wrote that the union of two sets is closed, which proves that $\overline A\cup\overline B$ is closed. It does not prove that it equals $\overline{A\cup B}$.
The problem with your second attempt is that you wrote that $$(A\cup B)'=A'\cup B'.$$
Do you have proof that this is so?
A: A standard way of showing two sets $S,T$ are equal to each other is by considering an element $s$ of $S$ and showing that $s$ belongs to $T$, and, conversely, by showing that every element $t$ in $T$ is contained in $S$.
You can use the fact here that the closure 
$\overline {A \cup B}$ equals $A\cup B $ together with the union of the respective limit points.  Now, consider an element $a$ in $\overline {A \cup B}$ and show it must belong to $\overline A \cup \overline B$.
Let me get you started: let $x$ be in $\overline {A \cup B}$ , then $x$ is either in $A$, in $B$ , or in the closure of the union (notice that $A\subseteq  \overline A$). If x is in $A$, then $x$ is in $A\cup B \subseteq \overline {A\cup B}$; similar if $x$ is in $B$. Now consider $y$ in $\overline {A \cup B}-A\cup B$...
