Prove that $x$ is a limit point of $A_1$ 
Let $A_m \subseteq \mathbb{R}^n$, $A_m \ne \emptyset$ and $A_{m+1} \subseteq A_m$. Suppose that $\bigcap\limits_{m=1}^{\infty} A_{m}=\emptyset$ and that $x \in \bigcap\limits_{m=1}^{\infty} \overline {A_{m}}$.
  Prove that $x$ is a limit point of $A_1$.

My attempt: I'm trying to prove that $Br^*(x)\cap A_1 \ne \emptyset , \forall r > 0$ or that $\exists$ {$x_n$} $\subseteq A_1\setminus${$x$} that $x_n \to x$. 
Let $x \in \bigcap\limits_{m=1}^{\infty} \overline {A_{m}}$ $\iff$ $x \in \overline {A_{m}}$, $\forall m \in \mathbb{N}$ $\iff Br(x)\cap A_m \ne \emptyset$, $\forall m \in \mathbb{N}$ and $\forall r>0$.
But since $x \notin$ $\bigcap\limits_{m=1}^{\infty} A_{m}$, so that $\exists$ $m \in \mathbb{N}$ that $x \notin A_m$.  By the hypothesis, we have that $x \notin A_{m+1}$, for this particular $m$. 
From this, how do I get that $x \in {A_1}'$? Any help to finish this proof would be appreciated. Thank you!

Hint 
Deduce that $x \in A_{m+1}'$. Since $A_{m+1} \subseteq A_1$ and $x \notin A_1$, you can then deduce that $x \in A_1'$.
