Completeness of the set of convergent sequences It's a problem from the book "Topology of Metric Spaces", written by Kumaresan:

"Show that the set $\textbf{c}$ of convergent sequences in the Normed Linear Space of all bounded real sequences under the sup norm $\|\|_\infty$ is complete. Hint: Enough to show that $\textbf{c}$ is closed. If x = $(x_n)$ is a limit point of $\textbf{c}$, it suffices to show that $x$ is Cauchy."

I got a little bit confused here... What I tried to do:
I know that to show that a metric space is complete, I have to show that every Cauchy sequence is convergent there. 
About the hint: I don't understand why it is enought to show that $\textbf{c}$ is closed, and what does it mean to say that $(x_n)$ is a limit point of $\textbf{c}$? Does that mean that there is a squence $(y_k) \in \textbf{c}$ such that $(y_k) \to (x_n)$? 
If $(y_k) \to (x_n)$, given $\epsilon > 0$, there are $n_0, k_0 \in \mathbb{N}$ such that
$$\|y_k-x_n\|_\infty<\epsilon \forall k,n \geq \max\{n_0, k_0\}=m_0$$
As we have that $(y_n) \in \textbf{c}$, we have that $(y_n) \to y \in \mathbb{R}$. Because of the uniqueness of the limit, $(x_n) \to y$, so $(x_n) \in \textbf{c}$, therefore, $\textbf{c}$ is closed. 
Now, let $(y_{n_k}) \in \textbf{c}$ be a Cauchy sequence. Well, $(y_{n_k})=(y_{n_1},y_{n_2},\dots,y_{n_j},\dots)$ is a sequence of sequences. Knowing that $(y_{n_k})$ is Cauchy's, we know that, given $\epsilon > 0, \exists j_0 \in \mathbb{N}$ such that $\forall m, p \geq j_0$
$$\|y_{n_m}-y_{n_p}\|_\infty < \epsilon$$
I don't know what to do from here. My idea is to show that, since every term $y_{n_j}$, of the sequence $(y_{n_k})$, belongs to $\textbf{c}$, therefore, every term $y_{n_j}$ is convergent, we have that $(y_{n_k})$ converges for a convergent sequence $(y_n)$. So, since $(y_n)$ is convergent, $(y_n) \in \textbf{c}$, we have that every Cauchy sequence in $\textbf{c}$ is convergent, meaning that $\textbf{c}$ is complete.
But I don't know how to do that, and I don't know if I understood the problem correctly, so I don't know if what I did before is correct...
I'd be really grateful if someone could give a hint, a solution, a coment about anything I wrote here! :) 
 A: 1)First notice that a closed subset of a complete space is complete. Another way of understanding closed sets is that a closed set contains all its limit points. Equivalently, a subset  $S$ is closed, if  every convergent sequence in $S$ has its
limit in $S$.


*An element $(x_n)$ is a limit point of a set $S$, if every neighborhood of $(x_n)$ intersects $S$. It is not always the case that if $(x_n)$ is a limit point, that there exists a sequence converging to $(x_n)$; this is true for 1st-countable spaces (which includes metric spaces).

3)Now, try to show c is closed by showing that if $(x_n)$ is a limit point of c, then $(x_n)$ is in $S$, or that, if you have a  convergent sequence in c, then its limit is also in c. The triangle inequality should help you there.
A general observation is that you can show the subspace $c$ is closed, by showing that its complement is open: consider a point $(x_n)$ in $\mathbb R^{\mathbb R} -c$ , i.e., a sequence of Reals that does not converge . Then show that there is an $\epsilon>0$ , so that the ball $B((x_n), \epsilon)$ (ball given in the $||.||_{\infty}$ metric ), so that no sequence in that ball converges. This should not be too hard; if a sequence $(x_n)$does not converge and $Sup|x_n-y_n| < \epsilon$ , then I think it is not too hard to show that $(y_n)$ does not converge either.
