Prove that $A$ is a closed set in $l_2$. Show that the set $\displaystyle A=\{x\in l_2:|x_n|\le\frac{1}{n}, n=1,2,\ldots\}$ is a closed set in $l_2$.
My attempt:
Suppose a sequence (of sequences) $(y^m)\subset A$ ($m$ is just a superscript, not an exponent) converges to a point $y\in l_2$. We will prove that $y\in A$. 
Let us denote the sequence $y^m$ by $(y^m_1, y^m_2, \ldots)$ and $y$ by $(y_1, y_2, \ldots)$. 
Let $\epsilon>0$. $\exists M\ge 1$ s.t. $\|y^m-y\|<\epsilon$ $\forall m\ge M$.
Suppose $y\notin A$. Then $\exists N\in\mathbb{N}$ s.t. $\displaystyle|y_N|>\frac{1}{N}$.
We note that $|y^m_N-y_N|\le\|y^m-y\|$.
So, $|y^m_N-y_N|<\epsilon$ $\forall m\ge M$
$\implies y^m_N-y_N\to 0$ as $m\to\infty$
$\displaystyle|y^m_N|\le\frac{1}{N}$ and $\displaystyle|y_N|>\frac{1}{N}$
So, $\displaystyle|y_N|-|y^m_N|>\frac{1}{N}-\frac{1}{N}=0$
By the reverse triangle inequality, $|y^m_N-y_N|\ge|y_N|-|y^m_N|>0$
How do I obtain a contradiction?
 A: Choose $\varepsilon=\lvert y_N\rvert-\frac{1}{N}\gt0$. Then $\lvert y_N\rvert-\frac{1}{N}\le\lvert y_N\rvert-\lvert y_N^m\rvert=\left\lvert\lvert y_N\rvert-\lvert y_N^m\rvert\right\rvert\le\lvert y_N-y_N^m\rvert$. (The last inequality is reverse triangle inequality.) So $\lvert y_N-y_N^m\rvert$ is not less than the positive number $\varepsilon=\lvert y_N\rvert-\frac{1}{N}$.
But, a proof of closedness could be much more simpler.
Fix a positive integer $j$. Consider the projection to $j$-th coordinate given by $\pi_i(x_1,x_2,\dots)=x_i$, which is continuous. Then the set $A_j=\{x\in l_2:\lvert x_j\rvert\le\frac{1}{j}\}$ is closed as a preimage of closed set: $A_j=\pi_j^{-1}\left(\left[-\frac{1}{j},\frac{1}{j}\right]\right)$. Your set $A$ is the intersection of the closed sets $A_j$.
A: Since $|y_N| > \dfrac 1N$ there is a positive number $c$ satisfying $|y_N| = \dfrac 1N + c$. Since $|y_N^m| \le \dfrac 1N$ you get
$$|y_N| - |y_N^m| \ge c$$ which contradicts $$|y_N| - |y_N^m| \le |y_N - y_N^m| < \epsilon$$
provided that $\epsilon > 0$ is chosen appropriately. In your proof you chose it too soon. It should be chosen after $c$ is specified.
Your proof could be done much simpler using a direct argument. Observe for all $n$ you have $|y_n| \le |y_n^m| + |y_n - y_n^m| \le \frac 1n + \|y - y^m\|_{\ell_2} \to \frac 1n$ as $m \to \infty$.
