Proving that $\{u_k\}_{k=1}^\infty$, $u_k=\left\{1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{k},0,0,\dots\right\}$, does not converge in a metric space 
Let $\ell$ be the set of sequences of real numbers where only a finite number of terms are different from zero $$\ell = \big\{\{x_n\}_{n=1}^\infty :x_i=0\text{ for all but a finite number of }i\text{-s}\big\}.$$ For $x=\{x_n\}$ and $y=\{y_n\}$ in $\ell$, define $$d(x,y)=\sup_{n\in\mathbb{N}}|x_n-y_n|.$$
   Let $u_k\in\ell$ be defined by $$u_k=\bigg\{1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{k},0,0,\dots\bigg\}.$$
  Show that $\{u_k\}_{k=1}^\infty$ is convergent, or show that $\{u_k\}_{k=1}^\infty$ is not convergent.

My strategy for showing that $\{u_k\}$ is not convergent is to  show that it converges to $a=\{1,\frac{1}{2},\frac{1}{3},\dots\}$ which is not contained in $\ell$. 
Let $\epsilon>0$ be given. Then we can choose $N=\lceil 1/\epsilon \rceil$, such that if $n\ge N$ then 
\begin{align}
d(u_n,a) &= d\left(\left\{1,\frac{1}{2},\dots,\frac{1}{n},0,0,\dots\right\}, \left\{1,\frac{1}{2},\frac{1}{3},\dots\right\}\right) \\
&= \frac{1}{n+1}<\frac{1}{n}\le \frac{1}{N} \le \epsilon.
\end{align}
Denoting the set of all bounded sequences of real numbers by $\ell^*$, i.e. $$\ell^* = \big\{\{x_n\}_{n=1}^\infty:x_i\in\mathbb{R} \text{ for all }i\in\mathbb{N} \big\},$$ it seems clear that $\ell\subset \ell^*$ and that $a\in\ell^*$. Also, since $\{u_k\}\rightarrow a\in\ell^*$ it cannot also converge to some other $b\in\ell^*$. Thus there is no way it can converge in $\ell$.
Does this make sense? Is this a valid strategy for showing that something does not converge in some metric space?
Any help would be greatly appreciated!
 A: Yes. The completion $\ell^*$ is a Hausdorff space, so the limit is unique. Since the limit lies in $\ell^*\setminus\ell$, there cannot be any other limit in $\ell$.
Needed to say, what you write as a definition of $\ell^*$ is not the space of bounded sequences, but the space of all sequences. Neither of these two is what you're looking for, because the completion $\ell^*$ contains all sequences $\{x_1,x_2,x_3,\dotsc\}$ such that $x_i\xrightarrow{i\to\infty}0$.
A: No, you haven't given $\ell^\star$ a topology so it doesn't make sense to say that $\{u_k\} \to a$ in $\ell^*$, and even then you would need to show that $\ell^*$ is separated to that the sequence can't converge to two limits at the same time.
Instead you can do a completely straightforward proof that it doesn't converge :
Let $a \in \ell$. You want to show that $(u_k)$ doesn't converge to $a$, so you need to give an $\epsilon > 0$ such that $d(u_k,a) > \epsilon$ for infinitely many $k$.
Since $a_i$ is nonzero for finitely many integers, there has to be some $i$ such that $a_i = 0$.
Then, for any $k \ge i$, $d(u_k,a) \ge |(u_k)_i - a_i| = \frac 1i$.
And so you are done by picking $\varepsilon = \frac 1i$
