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Can't find a flaw in that proof:

Induction by the length of derived series.

Base: if $[G, G]=e$ then the group is abelian...

Assume that statement is true for n-1.

We have group $G$ with the derived series length $n$, and a subgroup $H$.

$[G, G]=G'$ has derived series of length $n-1$, so $H \cap G'$ is finitely generated.

$H/(H\cap G')$ is a subgroup of finitely generated abelian subgroup $G/G'$. So $H$ is finitely generated.

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Well, where do you think the problem might be? If $H/(H\cap G^{\prime})$ is finitely generated then it isn't obvious that $H$ is. I would start by looking here, if I were you. – user1729 Feb 10 '12 at 14:48
@User1729: If he'd proven that $H\cap G'$ is finitely generated, then the finite generation of both $H/(H\cap G')$ and $H\cap G'$ would imply that $H$ is finitely generated: more generally, if $N$ and $G/N$ are finitely generated, then so is $G$: pick $g_1,\ldots,g_m$ whose images generated $G/N$, and $m_1,\ldots,m_k$ that generated $N$. Given $g\in G$, white $gN=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}N$, hence $g=g_{i_1}^{a_1}\cdots g_{i_{\ell}}^{a_{\ell}}n$ for some $n$, and now express $n$ in terms of $m_1,\ldots m_k$. The problem was elsewhere. – Arturo Magidin Feb 10 '12 at 17:36

You are implicitly assuming that $G'$ is finitely generated, but this need not be the case.

For example, consider the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$. This group is a semidirect product $\mathbb{Z}_2^\omega \rtimes \mathbb{Z}$, where $\mathbb{Z}_2^\omega$ is the direct sum of infinitely many copies of $\mathbb{Z}_2$, and $\mathbb{Z}$ acts on $\mathbb{Z}_2^\omega$ by translation. This group is clearly solvable, and it is has a standard two-element generating set. However, the commutator subgroup is a translation-invariant subgroup of $\mathbb{Z}_2^\omega$, and is therefore not finitely generated.

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In order to apply the induction hypothesis to $H\cap G'$, you need $G'$ to be finitely generated. Did you prove that the derived subgroup of a finitely generated solvable group is necessarily finitely generated?

HINT: The commutator subgroup of the free group of rank $2$ is free of infinite rank; what happens if you mod out by $[G',G']$? You get the free metabelian group of rank $2$. Prove that its commutator subgroup is free abelian of infinite rank.

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Another example is the Baumslag-Solitar group $G=\langle x,y\mid xyx^{-1}=y^2\rangle$ then the normal subgroup generated by $y$ is $A=Z[1/2]$ (dyadic rationals) an infinitely generated abelian group. In this case $A=G'$ the commutator subgroup.

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