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 $\hat{x} \in N_\alpha$.
We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$
Note that the Hölder norm is the "sum" of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder condition with parameter $\gamma \in (0,1)$ for the $k$-th derivatives. From the $C^k$ part of the Hölder norm and the completeness of $C^k(\overline U),$ we get a $u \in C^k(\overline U)$ with $\lim_{n\rightarrow\infty}u_n = u$ in $C^k(\overline U).$ Now let's look at the Hölder condition. Let's fix a multi-index $\alpha$ with $|\alpha| = k$ and write
$$
v := D^\alpha u\qquad and \qquad v_n := D^\alpha u_n
$$
for short. Then we have for fixed $x\neq y$
$$
\begin{align}
\frac{|v(x) - v(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) + v_m(x) - v_m(y) + v_m(y) - v(y)|}{|x-y|^\gamma} \\
& \leq \frac{|v(x) - v_m(x)|}{|x-y|^\gamma} + \frac{|v_m(x) - v_m(y)|}{|x-y|^\gamma} + \frac{|v_m(y) - v(y)|}{|x-y|^\gamma}
\end{align}
$$
for arbitrary $m \in \mathbb N.$ Here, the first and third summands on the r.h.s. can be made arbitrarily small by choosing $m$ large enough, since $v_m \rightarrow v$ uniformly, and the second summand on the r.h.s. is bounded independent of $m$ and $x\neq y$ since the sequence $u_n$ is Cauchy, hence bounded, in the Hölder norm. From all this we get a constant $M_\alpha$ such that
$$
\frac{|v(x) - v(y)|}{|x-y|^\gamma} \leq M_\alpha \qquad independent\ of\ x\neq y.
$$
This shows that we have
$$
u\in C^{k,\gamma}(\overline U).
$$
So far so good. Finally, we have to show that $u_n \rightarrow u$ in $C^{k,\gamma}(\overline U).$ From the construction of $u,$ we already have $u_n \rightarrow u$ in $C^k(\overline U),$ so we only have to show convergence in the Hölder seminorms of the $k$-th derivatives. We use the notation $\alpha,v,v_n$ from above. We have
$$
\begin{align}
\frac{|(v-v_m)(x) - (v-v_m)(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) - v(y) + v_m(y)|}{|x-y|^\gamma} \\
& = \frac{|(\lim_{k\rightarrow\infty}v_k(x)) - v_m(x) - (\lim_{k\rightarrow\infty}v_k(y)) + v_m(y)|}{|x-y|^\gamma} \\
& = \lim_{k\rightarrow\infty}\frac{|v_k(x) - v_m(x) - v_k(y) + v_m(y)|}{|x-y|^\gamma} \\
& = \lim_{k\rightarrow\infty}\frac{|(v_k - v_m)(x) - (v_k - v_m)(y)|}{|x-y|^\gamma}.
\end{align}
$$
Here, the r.h.s. can be made arbitrarily small independent of $x\neq y$ by choosing $m$ large enough, since the sequence $u_n$ is Cauchy in the Hölder norm. So we find $v_n \rightarrow v$ in the Hölder seminorm. Since there are only finitely many multi-indices $\alpha$ (or equivalently, $k$-th derivatives) to consider, we can conclude
$$
u_n \rightarrow u\qquad in\ C^{k,\gamma}(\overline U),
$$
as desired.
Best Answer
Define $$\Lambda: Y \to X: \alpha\mapsto \lim_{N\to \infty}\sum_{n=1}^N \alpha_ne_n.$$ We show $\Lambda$ is an isomorphism.
With this proposition we see $\|\Lambda^{-1}\|\leq C$.
Remark:
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.