One-relator Groups and Subgroups. I am currently working on "The Group Algebra of a torsion-free one-relator Group can be embedded in a Field." (Tekla and Jacques Lewin) In this Paper we find a corollary (the last one), which I am about to apply on something else. This corollary can only be aplied, if the group $G$ is a one-relator group. Since I am dealing with subgroups, I tried to figure out, when subgroups of one-relator groups are one-relator.

So let $G$ be a torsionfree one-relator group. 
Is then every subgroup $H<G$ one-relator?
 A: Your result holds for surface groups. That is, if $G$ is a surface group, so $G\cong\langle a_1, b_1, \ldots, a_g, b_g; [a_1, b_1]\cdots [a_g, b_g]\rangle$ or $G=\langle a, b, \ldots, c; a^2b^2\cdots c^2\rangle$, then every finite index subgroup of $G$ is the fundamental group of a surface group of the same or higher genus, and every infinite-index subgroup is free.
You say you are wondering when a subgroup of a one-relator group is one-relator. Rather unfortunately, you are in the bad setting - if your groups had torsion then things would be nicer. For example, there is a result of Steve Pride from 1977 which says that every two-generator subgroup group of a one-relator group with torsion is either a one-relator group with torsion or a free product of cyclic groups (the paper is entitled The two-generator subgroups of one-relator groups with torsion). The difference is that adding torsion means that your groups are hyperbolic, and actually there is a very nice result due to B.B.Newman, and enhanced by Gurevich, which tells you when a word represents the empty word. This result is called B.B.Newman's Spelling Theorem, or the Newman-Gurevich Spelling Theorem. The most complete statement I have been able to find of it is in the paper A spelling theorem for staggered generalized 2-complexes, with applications, by Howie and Pride.
Now, Pride's result is only applicable for two-generator subgroups. His paper has a counter-example for 3-generator subgroups, but this spits out a one-relator group without torsion, so isn't applicable here. Anyway, sticking with the hyperbolic theme, there is a result of Hill, Pride and Vella on the subgroups of "small cancellation" groups, which can be interpreted as saying that if your group has a generic relator then there are only finitely many conjugacy classes of two-generator subgroups. Basically, if you are going to find bad subgroups you are going to have to have more generators. In general, you might want to look up the work of Steve Pride from the 70s. It is good stuff, about generating groups and subgroups. Mostly about two generation though, and there is a lot of it!
Finally, I should tell you where there are lots of one-relator subgroups. Have you come across Magnus' method? There are two "forms" which this method takes. The first is using "staggered" presentations (I actually think that these may yield non-one-relator subgroups, but haven't tried enough examples to find out) and the second is using HNN-extensions. The point of this method is to write your one-relator group in terms of a one-relator group with a strictly shorter relator. In the HNN-method, we wish to write our group as an HNN-extension of such a group. For example, if $G=\langle a, b; aba^2b^2$ then rewrite $aba^2b^2$ as $ab^{-1}bab^{-1}ab^{-1}b^2=a^2b^{-1}ab$ using the map $a\mapsto ab^{-1}, b\mapsto b$ then $G=\langle a, b;a^2b^{-1}ab\rangle\cong\langle a, t; t^{-1}at=a^{-2}\rangle$. Therefore, this group is an HNN-extension of a cyclic group. This gives you an nice induction step. Magnus' used this idea to resolve the word problem for one-relator groups, and to prove that subsets of generators generate free groups (supposing your group is freely indecomposable). My point is, if you hang around the relator you will get one-relator groups, while if you don't take enough of it you get free groups! For references, Magnus' staggered presentation approach can be found in the book Combinatorial group theory by Magnus, Karrass and Solitar, while the HNN-method can be found in a paper of McCool and Schupp entitled On one relator groups and HNN-extensions. Another interesting reference is the article of Fine and Rosenberger entitled The Freiheitssatz and its extensions, but this is in a book and I haven't found it online.
Anyway, I will continue to try and come up with a finitely-generated counter example, but I though I would summarise some of my thoughts in the mean time. Thus a lengthy post which may or may not help you...
A: Consider the group $G=\langle a,b \mid aba^{-1}=b^2 \rangle$. This 1-relator torsion-free group contains a rank 1 infinitely generated abelian subgroup (dyadic rationals). This subgroup is not 1-relator. As for finitely generated subgroups, I am not sure, but, I suspect that the answer is still negative. 
Edit: Here is how to construct examples of finitely-generated subgroups which are not 1-relator. First, note that if $G$ is a 1-relator group, then $\operatorname{rank} H^2(G)\le 1$. In particular, the free product $A=\pi_1(S) \star \pi_1(S)$ (where $S$ is a closed oriented surface of genus $\ge 1$) is not a 1-relator group (as $\operatorname{rank} H^2(A) =2$). Now, many 1-relator groups $G$ contain such subgroups $A$ (with genus of $S$ at least 2). Once you have one such subgroup $A$, you can take its conjugate $gAg^{-1}$ via an element which does not normalize $A$ (assuming that such element exists). Then, I think, for large $n$ (at least if the group $G$ is torsion-free, hyperbolic and $A$ quasiconvex), the subgroup of $G$ generated by
$$
A, g^nA g^{-n}
$$
is isomorphic to the free product $A\star A$. Now, we just need a large supply of hyperbolic groups which contain quasiconvex surface subgroups $A$ and contain elements like $g$. I found such constructions in the paper
"One-ended subgroups of graphs of free groups with cyclic edge groups" by H.Wilton, http://arxiv.org/pdf/1102.2866v2.pdf.
There is further discussion of this in https://mathoverflow.net/questions/68132/fundamental-groups-of-surfaces. 
