Compactness in the space of all real null sequences I can‘t find a proof for this:
Let $c_{0}=\{ \left( x_{n}\right) _{n\in \mathbb{N}}\subseteq \mathbb{R}| \lim _{n\rightarrow \infty }x_{n}=0\}$ be the space of all real null sequences with the norm $\left\| x\right\|_\infty$.
Any subset $K\subseteq c_{0}$ is compact, iff $K$ is closed and bounded and $\lim _{n\rightarrow \infty }\sup _{x\in K}\left| x_{n}\right| =0$ is fulfilled.
The $\lim _{n\rightarrow \infty }\sup _{x\in K}\left| x_{n}\right| =0$ means:
The limit for $n\rightarrow \infty$ of the supremum of the set of the absolute values of the $n$-th elements of every sequence in $K$ exists and is equal to $0$.
Thanks for your help. :)
 A: It is easy to check that the metric space $c_0$ is complete. Now recall the compactness properties of subsets of complete metric spaces from [Eng] (see below). It follows that a subset $K$ of a complete metric space $(X,d)$ is compact iff $K$ is closed and totally bounded. The latter means that for each $\varepsilon>0$ the set $K$ has a finite $\varepsilon$-network $F$ that is a subset of $K$ such that for each point $x\in K$ there exists $y\in F$ such that $d(x,y)<\varepsilon$.
Now we are ready to prove the required claim.

Any subset $K\subseteq c_{0}$ is compact, iff $K$ is closed and bounded and $\lim _{n\rightarrow \infty }\sup _{x\in K}\left| x_{n}\right|=0$ is fulfilled.

($\Rightarrow$) Clearly, $K$ is closed and bounded. Suppose to the contrary that there exists $\varepsilon>0$ such that for each natural $N$ there exists $n>N$ such that $\sup _{x\in K}\left| x_{n}\right|\ge\varepsilon$. Let $F\subset K$ be a finite $\varepsilon/3$-network.  Since $F$ is finite and each $y\in F$ belongs to $c_0$, there exists a number $N$ such that $\|y\|_n<\varepsilon/3$ for each $y\in F$ and $n>N$. Let $(x_n)\in K$ be any sequence and $n>N$ be any number. Pick a point $y\in F$ such that $\|y-x\|<\varepsilon/3$. Then 
$$|x_n|=|(x_n-y_n)+y_n|\le |x_n-y_n|+|y_n|<\varepsilon/3+\varepsilon/3=2\varepsilon/3.$$
Thus $\sup _{x\in K}\left| x_{n}\right|\le 2\varepsilon/3<\varepsilon$, a contradiction. 
($\Leftarrow$) Since $K$ is closed, it suffices to show that given any $\varepsilon>0$ there exists a finite $\varepsilon$-network $F$ of $K$. Let $F$ be a maximal $\varepsilon$-disjoint subset of $K$. The latter means that $\|x-y\|\ge\varepsilon$ for each distinct points $x,y\in K$. The maximality of $F$ implies that it is a $\varepsilon$-network of $K$. We claim that $F$ is finite. Indeed, suppose to the contary that the set $F$ is infinite. Pick a number $N$ such that for each $n>N$, $\sup _{x\in K}\left| x_{n}\right|\le\varepsilon/4$. Since $\sup _{x\in F}\left| x_{1}\right|$ is finite, there exist an infinite subset $F_1$ of $F$ such that $|x_1-x’_1|\le\varepsilon/2$ for each $x,x’\in F$. Since $\sup _{x\in F}\left| x_{2}\right|$ is finite, there exist an infinite subset $F_2$ of $F_2$ such that $|x_2-x’_2|\le\varepsilon/2$ for each $x,x’\in F_1$. Continuing in this manner we obtain an infinite subset $F_N$ of $F$ such that $|x_i-x’_i|\le\varepsilon/2$ for each $x,x’\in F_N$ and each $1\le i\le N$. Since $|x_i|\le \varepsilon/4$ for each $i>N$, we see that $|x_i-x’_i|\le \varepsilon/2$ for each $x,x’\in F_1$ and each $i$, that is $\|x-x'\|\le \varepsilon/2$, a contradiction. 
References 
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.


