How to prove the limit points set of the union of some sets is closed? We use $d(A)$ to represent the limit points set of $A$.
$\{A_{\alpha},\alpha\in S\}$ is a family of sets,each $d(A_{\alpha})$ is closed.
I want to prove $d(\bigcup_{\alpha\in S}A_{\alpha})$ is closed.
I think I should prove $d(d(\bigcup_{\alpha\in S}A_{\alpha}))\subset d(\bigcup_{\alpha\in S}A_{\alpha}).$ So,but I don't know how to carry out.
Any help will be thanked.

For convenience let $X$ be the space, let $A=\bigcup_{\alpha\in S}A_\alpha$, and let $L=d(A)$; we want to show that $X\setminus L$ is open in $X$. 
Let $x\in X\setminus L$; then $x$ has an open nbhd $U_x$ such that $U_x\cap A\subseteq\{x\}$. If $U_x\cap A=\varnothing$, let $V_x=U_x$, and note that $V_x\subseteq X\setminus L$. 
If $U_x\cap A=\{x\}$, fix $\alpha\in S$ such that $x\in A_\alpha$, let $C=\operatorname{cl}\{x\}$, and let $D=C\setminus\{x\}$. Let $V_x=U_x\setminus d(A_\alpha)$; $d(A_\alpha)$ is closed, so $V_x$ is an open nbhd of $x$ contained in $U_x$. Moreover, $D\subseteq d(A_\alpha)$, so $V_x\cap D=\varnothing$. Suppose that $y\in V_x\setminus\{x\}$; then $y\notin C$, so $y$ has an open nbhd $W$ such that $x\notin W$, and it follows that $W\cap V_x$ is an open nbhd of $y$ disjoint from $A$ and hence that $y\notin L$. Thus, $V_x\subseteq X\setminus L$ in this case as well, and $X\setminus L$ is indeed open in $X$.
