Separable metric space with 0-dimensional kernel Suppose $X$ is a separable metric space, let $D(X)$ denote the Cantor-Bendixson derivative of $X$, and $D_\alpha(X)$ the $\alpha$-th derivative of $X$.
We denote $\operatorname{Ker}(X)$ the kernel of $X$, which is the least derived subspace of $X$ such that $D_\alpha(\operatorname{Ker}(X))=\operatorname{Ker}(X)$ for all $\alpha$. This implies that the kernel is a perfect space.
We say that a topological space is zero-dimensional if it is Hausdorff and has a basis of clopen sets.
Theorem: Suppose $X$ is a separable metric space, and $\operatorname{Ker}(X)$ is zero-dimensional, then $X$ is zero-dimensional.
The proof we were given is fairly constructive, and quite long and filled with details. However being the lazy person that I am, I was looking for a shorter proof.
My efforts came to this conclusion:
Suppose $\alpha=\gamma+\beta$ for $\gamma,\beta<\alpha$ and that for all $\beta<\alpha$ we have that if $D_\alpha(X)$ is zero-dimensional, then so is $X$, then $D_\alpha(X)$ being zero-dimensional implies $X$ is zero-dimensional.
Proof: Note that $D_\alpha(X) = D_\beta(D_\gamma(X))$, therefore $D_\gamma(X)$ is zero-dimensional, and therefore $X$ is.

This proof, however, does not hold for indecomposable ordinals (e.g. $\omega$, $\omega^\omega$, etc.) which makes it of very little use. I am certain that such argument can be made for indecomposable ordinals (or limit ordinals in general), I just can't find it.
Any help will be most appreciated.
Edit: It seems likely that the requirement that $X$ is a separable and metric space is redundant. In most likelihood it will be true for Hausdorff and (completely?) regular spaces. 
 A: (I believe that I have found a proof, I am not 100% sure about its correctness though.)
Theorem: Suppose $X$ is a metric space, then if $D_\alpha(X)$ is zero-dimensional, so is $X$.
Denote by $D_\alpha$ the $\alpha$-th derivative, and let $I_\alpha=D_\alpha\setminus D_{\alpha+1}$, that is the isolated points of $D_\alpha$.
Now suppose $D_\alpha$ is zero-dimensional, that is Hausdorff and has a basis of clopen sets. Denote such basis by $\mathcal B$.
We generate a clopen basis for $X$ by induction. Denote the topology $\tau$.
Take $B_0$ as the basis for $\tau^0$ formed as follow: $U\in B_0$ if and only if one of the following holds:


*

*$U\in\mathcal B$; for every $x\in U$, $x$ is not a limit point (in $X$) of the sequence from $X\setminus D_\alpha$;

*$U = V\cup\{x_n\in X\setminus D_\alpha\mid n<\omega\}$ where $x_n$ is a sequence whose limit is in $V$, for some $V\in\mathcal B$;

*$U=\{x\}$ for $x\notin D_\alpha$;

*$X\setminus U$ has one of the above properties.


Claim: $U\in B_0$ is clopen in the topology $\tau^0$.
Proof: The proof is trivial, since $X\setminus U\in B_0$.

Suppose $B_\gamma$ for $\gamma<\beta$ was defined and generates a zero-dimensional topology. We define a basis for $\tau^\beta$, $B_\beta$. Take $U\subseteq X$, then $U\in B_\beta$ if and only if one of the following holds:


*

*$U\in \bigcup_{\gamma<\beta} B_\gamma$ and for every $x\in U$, there is no sequence in $I_\beta$ such that $x$ is a limit point of it;

*$U = V\cup\{x_n\in \bigcup_{\gamma<\beta} I_\gamma\mid n<\omega\}$ for some $V\in \bigcup_{\gamma<\beta} B_\gamma$ and $\lim x_n\in V$;

*$U=\{x\}$ for $x\in I_\beta$;

*$X\setminus U$ has one of the above properties.


Claim: $U\in B_\beta$ is clopen in the topology $\tau^\beta$. (trivial as before)

Claim: $\tau=\tau^\alpha$.
Proof: Suppose $x_n\to x$ in $\tau$. If $x_n\in D_\alpha$ then $x\in D_\alpha$ and the convergence is same in $\tau^\alpha$. Suppose $x_n\notin D_\alpha$. If $x\in D_\alpha$ then for every clopen set containing $x$ we adjoined infinitely many $x_n$, so the convergence remains.
Otherwise, since $D_\beta$ are always closed we have that $x\in I_\beta$ such that $x_n\notin D_\beta$, in particular on the $\beta$-th level, $\tau^\beta$ was such that every set containing $x$ was adjoined with infinitely many $x_n$'s.
For the other direction we have that if $x_n\to x$ in $\tau^\alpha$ then once again we deal with the cases - if the sequence in $D_\alpha$ we are done, otherwise we have some $\beta$ in which $x\in I_\beta$. In particular $x_n\to x$ implies every open set with $x$ has infinitely many $x_n$'s, which was taken care of if and only if $x_n$ was a sequence of points which were removed before $D_\beta$ and $\tau$ has converging to $x$. In particular $x_n\to x$ in $\tau$.

We have that $\tau$ has $B_\alpha$ as a basis of clopen sets, as needed. (Separability and second-countability are retained by this process, and I believe the proof is easily generalized to Hausdorff spaces by replacing sequences with nets.)
A nice corollary is that if the kernel of $X$ is zero-dimensional, then $X$ is zero-dimensional.
