Theorem (suggested by I.V.Protasov). Every solvable Hausdorff topological group $G$ is topologically solvable in the sense that $G$ contains an increasing sequence of closed subgroups $\{1\}=G_0\subset G_1\subset\dots\subset G_n=G$ such that for every $i\le n$ the subgroup $G_{i-1}$ is normal in $G_i$ and the quotient group $G_i/G_{i-1}$ is abelian.
Proof: The group $G$, being solvable, contains an increasing sequence of subgroups $\{1\}=G_0\subset G_1\subset\dots\subset G_n=G$ such that for every $i\le n$ the subgroup $G_{i-1}$ is normal in $G_i$ and the quotient group $G_i/G_{i-1}$ is abelian.
For every $i\le n$ let $\bar G_i$ be the closure of the group $G_i$ in $G$. It can be shown that for every $i\le n$ the normality of the subgroup $G_{i-1}$ in $G_i$ implies the normality of the its closure $\bar G_{i-1}$ in the closure $\bar G_i$ of $G_i$. The quotient group $\bar G_i/\bar G_{i-1}$ is abelian since is contains a dense abelian group $G_i/(G_i\cap\bar G_{i-1})$ (which is a homomorphic image of the abelian group $G_i/G_{i-1}$). So, $G$ is topologically solvable.
Corollary. Each non-abelian solvable Hausdorff topological group contains a non-trivial closed normal subgroup and hence cannot be topologically simple.