Sequences in metric space $d({x_{n}}, {y_{n}}) = \sup\{|x_{n} – y_{n}|: n \in \mathbb{N}\}$[Edited] Let $c_{0}$ be the space of real-valued sequences $\{x_{n}\}$ which converge to zero, equipped with the metric $d(\{x_{n}\}, \{y_{n}\}) = \sup\{|x_{n} – y_{n}|: n \in \mathbb{N}\}$. Let $e_{k}$ denote the sequence in $c_{0}$ which is identically zero, except for the k-th entry which equals 1. 
(i) Prove that $\{e_{k}\}$ is a bounded sequence in $c_{0}$ (i.e., it takes values in a bounded set).
(ii)    Prove that $\{e_{k}\}$ has no convergent subsequence.
(iii)   Prove that the closed unit ball in $c_{0}$, B = $\{p \in c_{0} : d(0_{n}, p) \le 1\}$ is not compact. $0_{n}$ denotes a sequence of zeros.
(iv)    Prove that the metric space $(c_{0}, d)$ is complete.
Attempt:
(i)
For each $e_{k}$, its range is bounded, since $\sup_{m}\{e_{k}(m) – 0_{n}(m)\} = 1$ and so $d(e_{k}, 0_{n}) < 2$. 
(ii)
Here I am not sure that I understand this setting correctly. It seems that ${e_{k}} \rightarrow {0}$ in the usual metric, where 0 is a number, but ${e_{k}}$ fails to converge in the metric space $(c_{0}, d)$, since it has no convergent subsequence and a sequence is a subsequence of itself. 
But I still can not understand why ${e_{k}}$ fails to converge and here is my proof that it converges:
It is clear that should $e_{k}$ converge to something, it will converge to $0_{n}$. So we need: $\forall \epsilon$ $\exists N$, s.t. $n>N \implies \sup_{n}\{e_{k}(n) – 0_{m}(n)\} < \epsilon$. If we take $n > k$, then $\sup_{n}\{e_{k}(n) – 0_{m}(n)\} = 0 < \epsilon$, so $e_{k}$ converges and thus each subsequence of $e_{k}$ converges as well. 
Why is it wrong?
(iii). It seems to me that the only way to prove that the closed unit ball in this metric space is not compact is by using definition, i.e., finding an open cover without finite subcover. I tried several covers, but they do not work. E. g.:
Consider $A_{i}$. $A_{i} = \{p \in c_{0}: d(p, 0_{n}) \le 1 - \frac{1}{i}\}$. Then $\cup_{i}{A_{i}} \cup C$, where $C = \{p \in c_{0}: 1-\epsilon < d(0_{n}, p) < 1 + \epsilon\}$ covers B, but we can pick i, such that $\frac {1}{i} < \epsilon$, so i is finite and we have a finite subcover. It does not work.
EDIT:
So here $\{e_{k}\} = \{\{1, 0, 0, …\}, \{0, 1, 0, …\}, \{0, 0, 1, 0, …\}, …\}$.
(i)
$\{e_{k}\}$ is bounded, since its range is bounded. Let A = range of $\{e_{k}\}$. Then ${e_{k}} \in A$. Thus $d(e_{k}, 0_{n}) = 1$. So for any $e_{k}$, $d(e_{k}, 0_{n}) < 2$, where $e_{k} \in A$ and $0_{n} \in A$, meaning that $\{e_{k}\}$ is bounded.
(ii) and (iii) are answered below.
(iv)
Consider ${b_n}$, a sequence in $c_0$, which is Cauchy. Then $\forall \epsilon > 0$ $\exists N$, such that if $m \ge N$ and $n \ge N$, then $d(b_m, b_n) < \epsilon$. 
So the sequence ${b_n}$ converges to some sequence b. Since all $b_n$ and $b_m$, except of finitely many first elements, are identically zero, b will consist of zeros, except for finitely many elements at the beginning.
Thus $b \rightarrow 0$, so $b \in c_0$. Thus $c_0$ is complete.
Can you please verify proofs for (i) and (iv)?
 A: The sequence $e_k$ is convergent, just because it's stationnary, and also because it's an element of $c_0$ !
What I meant is that the sequence $(e_k)_k$ is not convergent : the "vectors" in $c_0$ are sequences, so a sequence of elements of $c_0$ is a sequence of sequences.
To prove that $(e_k)_k$ has no convergent subsequence, suppose $\ell$ is the limit of such a subsequence $(e_{\phi(k)})_k$. $\ell\in c_0$, so it is a sequence of limit $0$. So there is an integer $N$ such that for all $n\ge N$, $\left|\ell(n)\right|<\frac{1}{2}$.
But as $(e_{\phi(k)})_k$ converges to $\ell$, there must also be some $K\in\mathbb N$ such that for all $k\ge K$, $d(e_{\phi(k)},\ell)<\frac{1}{2}$.
Now this is a contradiction, as if $k$ is sufficiently large for $\phi(k)$ to be greater than $N$, you have to have 
$$1=\left|e_{\phi(k)}\right|\le \left|e_{\phi(k)}(\phi(k))-\ell(\phi(k))\right| + \left|\ell(\phi(k))\right|<\frac12+\frac12=1$$
So the hypothesis $(e_k)$ has a convergent subsequence has to be false.
A: To show that the closed unit ball $B$ is not compact.
Notation: $B_d(x,r)=\{y:d(x,y)<r\}.$
For brevity, for $k\in \mathbb N$ let $U_k=B_d(\{e_k\},\frac {1}{2}).$
For $k\ne k'$ we have $d(\{e_k\},\{e_{k'}\})=1$ so $U_k\cap U_{k'}=\emptyset.$
Let $V=\cup \{B_d(x,\frac {1}{2}): x\in B$ \ $\cup_{k\in \mathbb N} U_k\}.$
Let $C=\{U_k:k\in \mathbb N\}\cup \{V\}.$ Then $C$ is an infinite open cover of $B$ with no finite subcover. Because for each $k$, the only member of $C$ that contains $\{e_k\}$ is $U_k,$ and $\{U_k: k\in \mathbb N\}$ is an infinite set.  
