Prove that closure of A and A have the same accumulation points. (¯A=cl(A))
I have two attempts in mind.
(1) Show that the equality holds when A is closed, and when A is open.
(2) Show  A'⊆¯A' and ¯A'⊆ A'. 
My attempt:
(1) We can show that ¯A'=A' when A is closed.
When A is closed, (recalling the theorem: A is closed iff A=¯A) the derived set of A is the same as the derived set of ¯A (i.e. A'=¯A').
This means, ¯A and A have the same accumulation points when A is closed.
Now we are left to show that ¯A'=A' when A is open.
When A is open, let p∈A, then we can find an open ball B(p,r)⊂A. Since p is arbitrary, we know that every point of A has a neighbourhood (open ball) that contains at least one point of A. Hence, p is an accumulation point(i.e. every point in A is an accumulation point).
(I don't know what to do from here. How to show the accumulation points of ¯A is the same as that of A?)
(2) We can also show by definition: the intersection of the set of accumulation points of A and the set of accumulation points of B is equal to the set of accumulation points of the intersection of A and B (i.e. A'∩B'=(A∩B)').
So,
A'∩¯A'=(A∩¯A)' 
Since A⊆¯A, then
A∩¯A=A.
This gives
A'∩¯A'=A',
A'⊆¯A'.
Now we are left to show that ¯A'⊆A'.
(I have no idea how to show this)
Please advice.
 A: Definition:The closure of $A$ is defined as the common intersection of the family $F_A$  of all the closed sets that have $A $as a subset. Since the intersection of any family of closed sets is closed, therefore $\overline A$ is closed. And since $A$ is a subset of every member of $F_A,$ therefore $A\subset \cap F_A=\overline A.$
(i). Any member of the complement $\left(\overline A\right)^c$ of $\overline A$ belongs to an open set, namely, to $\left(\overline A\right)^c$, that is disjoint from $A.$ (ii). On the other hand if $p$ belongs to any open set $U$ that is disjoint from $A$ then  $U^c\in F_A$ so $U^c\supset \overline A$ so $p\not \in \overline A.$
Therefore by (i) and (ii) above, $p\not \in \overline A$ iff there exists an open set  that contains  $p$ and is disjoint from $A.$  Hence  $p'\in  \overline A$ iff every open set containing $p'$ has non-empty intersection with $A.$ (So we  could take that as an equivalent definition of  $\overline A.$)
$\bullet \;$.That is,  $p\in \overline A \iff p\in A'.$.. Therefore  $\overline A=A'.$
Now from the definition of the closure  of a set, if $B$ is closed then $B\in F_B$ so $\overline B=B.$ Applying  this, and $\bullet\;$, with $B=\overline A,$ we have   $$\left(\overline A\right)'=B'=\overline B=B=\overline A=A'.$$
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