Show $A =\{x\in \ell_2 : |x_n|\le\frac{1}{n}\},n=1,2,...$ is a closed set in $\ell_2$ but $B=\{x\in \ell_2 :|x_n|< \frac{1}{n}\}$ is not an open set. Show that the set $A =\{x\in \ell_2 : |x_n|\le\frac{1}{n},n=1,2,...\}$ is a closed set in $\ell_2$ but that $B=\{x\in \ell_2 :|x_n|&lt; \frac{1}{n}\}$ is not an open set.

I think if I show each $A_n=\{x\in\ell_2,|x_n|\le\frac{1}{n}\}$ is closed, I can get $A=\displaystyle\bigcap_{n}A_n$ that is closed. Can I do that? However, I didn't know how to prove $A_n=\{x\in\ell_2,|x_n|\le\frac{1}{n}\}$ is closed. 

Are there any other ways to prove this?
As @SangchulLee comment convergence in $\ell_2$ implies coordinate wise converge, therefore $x_n\overset{\ell_2}\rightarrow x$ implies that for all $k:x_n^{(k)}$ converges to $x^k$ in $\mathbb R$.

Now the point is that $|{x}|\leq \frac 1 n$ is a closed set, therefore if $|x_n^{(k)}|\leq \frac 1 k$ and it converges the limit element will also satisfy $|x^{(k)}|\leq\frac 1 k$ and so it effectively lies in $A$.

For the second case you have that $x_n^{(k)}&lt;\frac 1 k$ which is an open condition so that the limit can escape. For example you can take the sequence: $$x_n=\begin{cases} 1-\frac 1 n,\quad k=1\\0\end{cases}$$

this is clearly in $B$ bu the limit is the sequence $(1,0,0,\dotsc)$ which does not lie in $B$.
You could try showing it's the inverse image of a closed set under a continuous map.