Here is a sketch. I will use all of the same notations you are using (You should definitely check it for mistakes)
For each $n$, then sequence $\xi_n=(x^{(n)}_k)_{k\geq 0}$ is a convergent sequence (because it is in $c$), so let us say that $\xi_n=(x^{(n)}_k)_{k\geq 0}$ converges to $u_n \in \mathbb{C}$.
You can check that $|u_m-u_n|\leq ||\xi_m-\xi_n||_{\infty}$ for all $m,n \in \mathbb{N}$, so since $(\xi_n)_{n\geq 0}$ is a Cauchy sequence in $c$, it follows that $(u_n)_{n \geq 0}$ is a Cauchy sequence in $\mathbb{C}$, so there exists $u \in \mathbb{C}$ such that $\lim_{n\to \infty}u_n=u$.
I claim that the sequence $\xi=(x_k)_{k\geq 0}$ converges to $u$. The reason is that for any $n$ we have $$|x_k-u| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-u_n|+|u_n-u| $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-u_n|+|u_n-u|$$Let $\epsilon>0$. Choose $n_0 \in \mathbb{N}$ such that $|u_{n_0}-u|<\frac{\epsilon}{3}$ and $||\xi-\xi_{n_0}||<\frac{\epsilon}{3}$. Choose $K \in \mathbb{N}$ such that $|x_k^{(n_0)}-u_{n_0}|<\frac{\epsilon}{3}$ whenever $k \geq K$. Then for $k \geq K$, all three terms on the right hand side of the above equation are less than $\frac{\epsilon}{3}$, and thus $|x_k-u|<\epsilon$.
Thus $x_k \to u$. Therefore the sequence $\xi=(x_k)_{k\geq 0}$ converges, so $\xi \in c$.
Edit: (Alternative solution) Actually, I just realized that we don't really care about the limit of $\xi$, we only care that it converges. So it suffices to prove that it is Cauchy. To do this, you can just say that for any $n$ $$|x_k-x_l| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-x^{(n)}_l|+|x^{(n)}_l-x_l|$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-x^{(n)}_l|+||\xi-\xi_n||_{\infty}$$ and then argue as before.
Define $$\Lambda: Y \to X: \alpha\mapsto \lim_{N\to \infty}\sum_{n=1}^N \alpha_ne_n.$$ We show $\Lambda$ is an isomorphism.
- By the definition of $Y$-norm, $\Lambda$ is contractive, hence continuous.
- $\Lambda$ is onto, since {$e_n$} is a Schauder basis.
- $\Lambda$ is injective. Assume $\Lambda \alpha =0$, that is $\sum_{n=1}^\infty \alpha_ne_n=0$. Again, since {$e_n$} is a Schauder basis , the $\omega$-linear independence of $\{e_n\}$ gives $\alpha_n=0$ for all $n$.
- It remains to show $\Lambda^{-1}$ is bounded. This is the key step. The following proposition is an equivalent definition of Schauder basis.
Proposition: $\{e_n\}$ is Schauder basis if and only if it is $\omega$-linear independence basis, and there exists a constant $C>0$, such that $\|P_N\|\leq C$. Here $P_N$ is the projection $$P_N:X\to X: x=\sum_{n=1}^\infty \alpha_n e_n \mapsto\sum_{n=1}^N\alpha_ne_n.$$
With this proposition we see $\|\Lambda^{-1}\|\leq C$.
Remark:
- above proposition is not trivial. It can be found in text books and I will add references later.
- You can't use Banach inverse operator theorem to replace the fourth step, since we don't know $Y$ is Banach yet.
One reference is Topics in Banach spaces theory, Proposition 1.1.9. As David Mitra said, it this proposition may equivalent with that $X$ is isometric to $Y$. I'm sorry for not checking it when I post this answer. Anyway, this book will provide a self-contained proof to your question.
Best Answer
Suppose $x_n$ is Cauchy. We need to first produce a candidate limit, and then show that this limit is in $N_\alpha$.
(I am assuming that you have shown the various properties of the norm, in particular, $\|x+y\| \le \|x\|+ \|y\|$.)
Then for all $\epsilon>0$ there exists $N$ such that if $n,m \ge N$ then, for all $p$, $\sum_{k=1}^p |[x_n]_k-[x_m]_k| \le \epsilon p^\alpha$. In particular, for a fixed $p$, $|[x_n]_p-[x_m]_p| \le \epsilon p^\alpha$. Hence $[x_n]_k$ is Cauchy and converges to a limit. Let $[\hat{x}]_k = \lim_{n \to \infty} [x_n]_k$. The sequence $\hat{x}$ is our candidate limit.
Now we must show that $\hat{x} \in N_{\alpha}$.
Now fix $\epsilon =1$, and choose $N$ such that if $m,n \ge N$, then $\sum_{k=1}^p |[x_n]_k-[x_m]_k| \le p^\alpha$. If we fix $p$ and then take the limit as $n \to \infty$, and set $m=N$, we have $\sum_{k=1}^p |[\hat{x}]_k-[x_N]_k| \le p^\alpha$. If we let $[y]_k = [\hat{x}]_k-[x_N]_k$, the above shows that $y \in N_\alpha$.
Since $\|x_N + y\| \le \|x_N\|+ \|y\|$, we have $x_N + y \in N_\alpha$, and since $[\hat{x}]_k = [x_N]_k + [y]_k$, we see that $\|\hat{x} \| \le \|x_N\|+ \|y\| < \infty$, hence $\hat{x} \in N_\alpha$.