This follows from the following theorem, which is a common ingredient in the proof of the structure theorem for finitely generated abelian groups:
Theorem. Let $r\gt 0$ be a positive integer, and let $H$ be a subgroup of $\mathbb{Z}^r$. Then there exists a basis $a_1,\ldots,a_r$ of $\mathbb{Z}^r$, an integer $d$, $0\leq d\leq r$, and positive integers $m_1,\ldots,m_d$ such that $m_1|m_2$, $m_2|m_3,\ldots,m_{d-1}|m_d$ such that $m_1a_1,\ldots,m_da_d$ is a basis for $H$. In particular, $H$ is free and finitely generated.
You can see a proof of this in this previous answer.
To see how this proves the result, suppose that $G$ is abelian and finitely generated by $g_1,\ldots,g_r$. Let $H$ be a subgroup of $G$. There is a surjection $\mathbb{Z}^r\to G$ given by mapping the standard basis vector $\mathbf{e}_i$ to $g_i$; the subgroup $H$ corresponds to a subgroup $\mathcal{H}$ of $\mathbb{Z}^r$ by the isomorphism theorems. By the Theorem, $\mathcal{H}$ is finitely generated, and hence its image, $H$, is also finitely generated (generated by the images of the generators of $\mathcal{H}$).
That's a very strangely worded question, but I think it's probably meant to be a fill-in the blank multiple choice question which would more intelligibly be stated as:
Every subgroup of $(\mathbb{Q}, +)$ is ____________________.
A) cyclic and finitely generated, but not necessarily abelian or normal.
B) cyclic and abelian, but not necessarily finitely-generated or normal.
C) abelian and normal, but not necessarily cyclic or finitely-generated.
D) finitely generated and normal, but not necessarily cyclic or abelian.
Since $(\mathbb{Q}, +)$ is an abelian group, every subgroup is obviously abelian and normal, which rules out A, B, and D, so the answer is C. Of course you want to actually be able to put your hands on a non-finitely generated subgroup of $(\mathbb{Q}, +)$, so as anon says in the comments, consider the additive group of $\mathbb{Z}[\frac{1}{2}]$, i.e. the subgroup of $(\mathbb{Q}, +)$ consisting of elements where the denominator is a power of 2.
(Note that a finitely-generated subgroup of $(\mathbb{Q}, +)$ has an upper bound on its denominators, and is in fact cyclic.)
Best Answer
Hint 1. Given any two elements $a,b\in\mathbb{Q}$, can you find an element $r\in\mathbb{Q}$ such that $a=mr$ and $b=nr$ for some integers $m$ and $n$? If so, then $\langle a,b\rangle\subseteq \langle r\rangle$; what do we know about subgroups of cyclic groups.
Hint 2. Is there a finitely generated subgroup of $\mathbb{Q}\times\mathbb{Q}$ that is not cyclic?