The point of this section of Stroppel's book is to show, ultimately, that nothing new happens. Stroppel shows that each term in the Hausdorff derived series is nothing other than the closure of the same term in the usual derived series. A topological group is Hausdorff-solvable if and only if it is solvable, and the solvable height equals the Hausdorff-solvable height. In a sense, you can't construct interesting examples. :-)
One thing that you can do is make an example of a topological group whose commutator subgroup isn't closed. I cheated with Google to find this, but here goes anyway. There exists a finite group $G_n$ which is 2-step nilpotent and such that the commutator subgroup requires a product of $n$ commutators. Namely, take a central extension of a $2n$-dimensional vector space $V$ over an odd finite field by its exterior square $\Lambda^2 V$, such that the commutator of $a,b \in V$ is $a \wedge b \in \Lambda^2 V$. The point is that you need $k$ commutators to reach a tensor in $\Lambda^2 V$ of rank $k$. Now let $G$ be the product of all $G_n$ in the product topology. The algebraic commutator subgroup of $G$ isn't closed, because it does not include elements in the closed commutator subgroup whose commutator length in $G_n$ is unbounded as $n \to \infty$. Amazingly, this group $G$ is even compact.
The implication for a Galois algebraic field extension, say, is as follows. The Galois group $G$ of such a field extension is a topological group, in fact a profinite group. You might have wondered if the algebraic field extension is "solvable" in the group-theoretic sense, but without leading to solvability by radicals. Happily, it doesn't happen, because what you should do is replace the solvable series of $G$ by the closed solvable series.