- Start with a sequence in $c_0$, our goal is to show that its limit $\omega\in\ell^\infty$ is also in $c_0$. This limit exists by completeness of $\ell^\infty$
Incorrect. To prove that a set is closed, we start with a convergent sequence in that set, and show that the limit is contained in the set. The existence of limit is assumed, not obtained from completeness.
"Quick remark" is totally wrong. Take $x_n = (1+1/n,0,0,0,\dots)$. Then the limit of $x_n$ is $(1,0,0,\dots)$, not the zero sequence.
I don't know where you got $\omega_n^{(k)}$ from, or what it means; consequently I can't make any sense from your items 3 and following.
Advice: since you are getting confused by "sequence of sequences" business, change the notation to functions. A sequence of real numbers if just a function $f:\mathbb N\to\mathbb R$. So, you have a sequence of functions $f_n:\mathbb N\to\mathbb R$. They converge to some function $g$ in the sense that $\|f_n-g\|_\infty\to 0$. The goal is to prove that the function $g$ satisfies $\lim_{k\to\infty} g(k)=0$.
By the definition of limit, we must do the following: given $\epsilon>0$, find $K$ such that $|g(k) |<\epsilon$ whenever $k\ge K$. To this end, we
- find $N$ such that $\|f_n-g\|_\infty<\epsilon/2$ whenever $n\ge N$.
- find $K$ such that $|f_N(k)|<\epsilon/2$ whenever $k\ge K$. (This is possible since $f_N\in c_0$)
- observe that $|g(k)|\le |f_N(k)|+|f_N(k)-g(k)| <\epsilon$ whenever $k\ge K$, as required.
To see that a subspace of a metric space is closed, you only need to take any convergent sequence in the subspace and prove that the limit belongs to the subspace. In this case take a sequence $\left( x^{(n)}\right)_{n=1}^{\infty} \subset c_0$, such that $x^{(n)} \underset{n \to \infty}{\longrightarrow} x \in \ell^{\infty}$. Since each $x^{(n)} \in c_0$, we have
\begin{equation}
\lim_{j \to \infty } x_j^{(n)} = 0 \ \ \text{ for each $n \in \mathbb{N}$ } \ \ \ \ \ (1)
\end{equation}
Also, for every $\varepsilon >0$, there exist $n=N(\varepsilon) \in \mathbb{N}$, such if $n > N$, then
$$
\|x^{(n)}- x \|< \varepsilon
$$
So, for each $j \in \mathbb{N}$, if $n>N$
$$
|x^{(n)}_j - x_j| \leq \sup_{j \in \mathbb{N}} |x^{(n)}_j - x_j| = \|x^{(n)}- x \|< \varepsilon
$$
that is, for all $j \in \mathbb{N}$, $x_j^{(n)} \underset{n \to \infty}{\longrightarrow} x_j \in \ell^{\infty} \ \ (2)$, an it converges uniformly ${(3)}$. Then
$$
\lim_{j\to \infty} x_j \overset{(2)}{=} \lim_{j\to \infty} \left( \lim_{n \to \infty} x^{(n)}_j\right) \overset{(3)}{=}\lim_{n\to \infty}\left(\lim_{j \to \infty} x^{(n)}_j\right) \overset{(1)}{=} \lim_{n \to \infty} 0 = 0
$$
that is , $x=\left( x_j\right)_{j=1}^{\infty} \in c_0$, so we conclude that $c_0$ is a closed subspace of $\ell^{\infty}$.
Best Answer
Yes, the author of that post only tries to prove that $c_0$ is a closed subset of $\ell^\infty$, in spite of the title of the post. Actually, the statement is “Prove that $c_0$ is closed in $\ell^\infty$.” So, yes, you still have to prove that it is a subspace (which is very easy, by the way).