I am very doubtful that the proof of the Cantor Bendixson Theorem using Cantor Bendixson derivatives need needs to use the fact that every subsequence of $\{\omega, \omega^\omega, ...\}$ has a larger element in the sequence, which is not true as you pointed out.
A usual proof of the Cantor Bendixson Theorem using derivative goes as follows: For any set $X \subset \mathbb{R}$, let $X'$ be the set of limit points of $X$. Define by transfinite recursion on the the ordinals,
$X_0 = X$
$X_{\alpha + 1} = X_\alpha'$
$X_\lambda = \bigcap_{\gamma < \lambda} X_\gamma$ when $\lambda$ is a limit ordinals.
The claim is that there exists a countable ordinals $\beta$ (i.e. $\beta < \omega_1$) such that $X_\beta = X_{\beta + 1}$.
To prove this, fix $(U_n)_{n \in \mathbb{N}}$ to be a countable basis for $\mathbb{R}$. For any closed set $F$, define $N(F) = \{n \in \omega : U_n \cap F \neq \emptyset\}$. Since $F$ is closed, $\mathbb{R} - F = \bigcup_{n \notin N(F)} U_n$. Thus for any two closed sets $F \neq E$, $N(F) \neq N(E)$. Moreover, if $F \subseteq E$, then $N(F) \subseteq N(E)$. Hence it has been shown that if $F \subsetneq E$, then $N(F) \subsetneq N(E)$.
Now applying this to the sequence of closed sets, $(X_\alpha)$. For any $\gamma < \alpha$ such that $X_\alpha \subsetneq X_\gamma$, one has $N(X_\alpha) \subsetneq N(X_\gamma)$. Since $N(X_\alpha)$ are subsets of $\mathbb{N}$ and $\mathbb{N}$ is countable, you can not have a uncountable $\beta$ such that for all $\eta < \xi < \beta$, $N(X_\xi) \subsetneq N(X_\eta)$.
Hence it has been shown that there exists a countable $\beta$ such that $X_\beta = X_{\beta + 1}$. It is clear that $X_\beta$ has no isolated points. Thus $X_\beta$ is your perfect kernel.
There is a easier proof of Cantor Bendixson that does not use ordinals. Note that a condensation point is a point $x$ such that every neighborhood of $X$ is uncountable. Let $Z$ denote the set of condensation points of $X$. The claim is that $X = Z \cup C$ where $C$ is countable and $Z$ is perfect. Let $C = X - Z$. Let $U_n$ be a countable basis for $X$ (i.e. take a countable basis for $\mathbb{R}$ and intersect it with your closed set $X$). By definition of $Z$, $C$ is the union of all $U_n$ such that $U_n$ is countable. A countable union of countable set is countable, hence, $C$ is countable. $Z$ is perfect because for any $x \in X$, take any open set $x \in V$. By definition of $x \in Z$, $V$ contains uncountable many points. $C = X - Z$ is countable, so $V$ actually contains uncountable many points of $Z$. Thus $X = Z \cup C$, where $Z$ is perfect and $C$ is countable.
It seems like a more obvious proof is to avoid the corollary and use the theorem directly:
If $x$ is not an isolated point, then $X\setminus\{x\}$ is dense and open in $X$.
If $S=\{s_1,s_2,\dots\}$ is countable, define $V_n=X\setminus \{s_n\}$, which are dense and open since $X$ has no isolated points. Then $\emptyset = \bigcap V_n\cap\bigcap U_n$ is the the countable intersection of dense open sets.
This is essentially the dual of your proof.
Best Answer
Start with $\{0\}\cup \{\frac{1}{n+1}\mid n\in \omega\}$. Now for each $n\in \omega$, add points in the interval $(\frac{1}{n+1},\frac{1}{n})$ limiting to $\frac{1}{n+1}$, to ensure that $\frac{1}{n+1}$ is isolated in $X^{(n)}$, but not in $X^{(n-1)}$. (In other words, you want to make $X\cap [\frac{1}{n+1},\frac{1}{n})$ into a space of Cantor-Bendixson rank $n+1$, which you say you know how to do in your question.)
Then $\bigcap_{n\in \omega} X^{(n)} = \{0\}$, so $X$ has Cantor-Bendixson rank $\omega+1$.