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.