Prove that the derived set of $A \subseteq \Bbb R$ is closed For $A \subseteq \Bbb R$, the derived set of $A$, denoted by $A'$, is the set of all limit points of $A$.

Theorem: For $A \subseteq \Bbb R$, $A'$ is closed.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:
Assume that $a \in \Bbb R$  is a limit point of $A'$ . Then for all $\dfrac{\delta}{2}>0$ , there exists $y \in A'$  such that $y \neq a$  and $|y-a|<\dfrac{\delta}{2}$. Let $\delta' = \dfrac{|y-a|}{2}>0$.
$y \in A'$ $\implies$ there exists $x \in A$  such that $x \neq y$  and $|x-y|< \delta'$.
It follows that $|x-y| < \delta' = \dfrac{|y-a|}{2} < |y-a|<\dfrac{\delta}{2}$ and $|y-a| - |x-y| > |x-y|>0$.
As a result,


*

*$|x-a| = |(x-y)+(y-a)| \le |x-y| +|y-a|<\dfrac{\delta}{2}+\dfrac{\delta}{2}<\delta$.

*$|x-a| = |(x-y)+(y-a)| \ge ||y-a| -|x-y||> |x-y| >0$. Thus $x \neq a$.
Hence, for all $\delta >0$, there exists $x \in A$ such that $x \neq a$ and $|x-a| < \delta$. So $a$ is a limit point of $A$ and thus $a\in A'$. Finally, $A'$ is closed.
 A: Another approach is to show the complement is open. Let $x\in (A')^c.$ Then, $x$ is not a limit point of $A$, which means there is an open neighborhood $x\in U(x)$, such that either $U(x)\cap A=\emptyset$ or $U(x)\cap A=\{x\}$. In either case, we have $x\in (A')^c.$ 
Now, if $y\in U(x)$, then there is an open neighborhood $y\in  U(y)\subseteq U(x)$, so $U(y)\cap A=\emptyset,$ from which it follows that  $y\in (A')^c$ and therefore that $U(x)\subseteq (A')^c$ so $(A)^c$ is open.
A: It is well known that : $\overline{X} = X \cup X'$. So a set $X$ is closed if and only if $X' \subseteq X$. 
Let $a \in (X')'$ and $\epsilon >0$, it has to $(a- \epsilon,a+ \epsilon ) \cap (X'-\{a\}) \neq \emptyset $ so exists $x_1 \in (a- \epsilon,a+ \epsilon ) \cap X',x_1 \neq a$.
Suppose, without loss of generality, that $x_1<a$ and let´s put $ \delta = min \{a-x_1,x_1-a+ \epsilon \} >0$. Since $x_1 \in X' : \exists$ $x_2 \in (x_1- \delta,x_1+ \delta ) \cap (X-\{x_1\}) \neq \emptyset $. 
It's easy to see that $(x_1 - \delta,x_1 +\delta) \subseteq (a- \epsilon,a) \subseteq (a- \epsilon,a+ \epsilon)$ so $x_2 \in (a- \epsilon,a+ \epsilon ) \cap X$.
Furthermore if $x_2=a$ then $x_2=a<a$ imposible so $(a- \epsilon,a+ \epsilon ) \cap (X-\{a\}) \neq \emptyset $. Finally $a \in X'$ and by the observation above the result is concludes.
