The closure of $c_{00}$ is $c_{0}$ in $\ell^\infty$ let
$x=(x(1),x(2),\ldots,x(n),\ldots)\in c_0$
So for any $\varepsilon>0$ there exists a $n_0\in\mathbb{N}$ such that $|x(n)|<\varepsilon_2$ for all $n\geq n_0$.
Now for all $n\geq n_0$ ,$||x_n−x||_\infty=\sup{|x(m)|}_{m\geq n_0}\leq \varepsilon_2<\varepsilon$
i have a doubt that regarding the last line. What makes it possible?
 A: I presume that $x_n=(x(1),\ldots,x(n),0,\ldots)$
In that case $x-x_n=(0,\ldots,0,x(n+1),x(n+2),\ldots)$
Hence $||x_n−x||_\infty=\sup_{m\geq n+1}|x(m)|$.
If $n\geq n_0$, $||x_n−x||_\infty=\sup_{m\geq n+1}|x(m)|\leq \sup_{m\geq n_0+1}|x(m)|\leq \epsilon$
A: I assume what you want to show is that the closure with respect to the sup-norm, of the space of all sequences that are eventually zero ($c_{00}$) is the space of all sequence that tend to $0$ that is $c_0$.
To show this  some sequence $x=(x(1), x(2),  \dots) \in c_0$ is fixed and one needs to show that for every $\epsilon >0$ there is a $y \in c_{00}$ with 
$\|  x- y \| < \epsilon$. 
To do this one observes that for any $\epsilon_2$ there is some $n_0$ such that $|x_n| < \epsilon_2 $ for all $n > n_0$. Now let $x_{n_0}= (x(1), x(2), \dots , x(n_0), 0, 0, \dots ) \in c_{00}$. 
Then $x-x_{n_0}= (0, \dots , 0, x(n_0+1),x(n_0+2), \dots ))$. 
Now $x(n)< \epsilon_2$ for all $n > n_0$. So $\| x- x_{n_0}\| \le \epsilon_2$. 
Thus given any $\epsilon > 0$ choosing some $0 < \epsilon_2 < \epsilon$ and proceeding as above you get some $x_{n_0}$ in $c_{00}$ such that $\| x- x_{n_0}\| \le \epsilon_2 < \epsilon$. 
