So the proof goes as follows, asumming $\ell_\infty$ metric:
Let $x^{(k)} \in c_0$ be a sequence converging to $\omega \in \ell^{\infty}$. Let $\epsilon > 0$ and $N \in \mathbb{N}$ such that $\sup|x_n^{(k)} – \omega_n| < \epsilon/2 \ \forall k > N$. For each $k$ choose $M \in \mathbb{N}$ such that $|x_n^{(k)}| \leq \epsilon/2 \ \forall n \geq M$. This means we can write $$|\omega_n| \leq |\omega_n – x_n^{(k)}| + |x_n^{(k)}| \leq \epsilon$$ for $n > M$ and $k > N$. This means the sequence $\omega_n$ converges to $0$ and $\omega \in c_0$.
First of all the notation is somewhat confusing. $\omega$ is the limit $x^{(k)}$ converges to, so $\omega$ must be a scalar. However, $\omega_n$ denotes a sequence. Is it such that each $x^{(k)}$ converges to an element in $\omega_n$? If so wouldn't $\omega_k$ be a more intuitive notation?
Also, we write $|x_n^{(k)}| \leq \epsilon/2 $ but which is not true for all $\epsilon > 0$ (unless we know the sequence converges to zero). But the last inequality must hold for all $\epsilon > 0 $ for us to conclude that $\omega_n$ converges to a sequence.
Best Answer
No. $x^k \in c_0$ is a sequence and the sequence of sequences $(x^k)_{k \in \mathbb N} \subset c_0$ converges to the sequence $\omega \in \ell^{\infty}$. In the second part of the proof you look at the scalar entries $\omega_n \in \mathbb C$ of the sequence $\omega$ (and similar the entries of the sequence $x^k$).
Furthermore, since $x^k \in c_0$, we have $x_n^k \to 0$ and thus one can find an $\varepsilon > 0$ such that $|x_n^k| \le \frac{\varepsilon}{2}$.