We want to prove by induction on $n$ that every subgroup of an abelian group generated by $n$ elements is finitely-generated. If $n=1$, it is clearly true. From now on, suppose it is true for $n$.
Let $G= \langle a_1, \dots, a_{n+1} \rangle$ be an abelian group, $H \leq G$ be a subgroup and $\rho : G \to G/ \langle a_{n+1} \rangle$ be the quotient map. Notice that the group $G/ \langle a_{n+1} \rangle$ is generated by $\{ \rho(a_1) , \ldots, \rho(a_n) \}$, so by our induction hypothesis, the subgroup $\overline{H}:= \rho(H)$ is finitely generated: let $X=\{h_1,\dots, h_m\} \subset H$ be such that $\rho(X)$ generates $\overline{H}$.
On the other hand $H \cap \langle a_{n+1} \rangle$ is a cyclic group, say generated by $h_{m+1} \in H$. Now, we want to prove that $Y= \{h_1, \dots, h_m,h_{m+1} \}$ generates $H$.
Let $h \in H$. Of course, there exists a word $w \in \langle h_1,\dots, h_m \rangle$ such that $\rho(w)=\rho(h)$. Therefore, $h=w+k$ for some $k \in \mathrm{ker}(\rho)= \langle a_{n+1} \rangle$. Furthermore, $k=h-w \in H$ so $k=p \cdot h_{m+1}$ for some $p \in \mathbb{Z}$ since $\langle a_{n+1} \rangle \cap H = \langle h_{m+1} \rangle$. Finally, $$h=w+p \cdot h_{m+1} \in \langle h_1, \ldots, h_m,h_{m+1} \rangle = \langle Y \rangle,$$
so $Y$ generates $H$.
We say that G is finitely generated if a basis for G is finite.
This isn't the usual definition of finitely generated. Usually, $G$ is said to be finitely generated if there exist $e_1, ... , e_n \in G$ such that every element of $G$ can be expressed as
$$g = c_1e_1 + \cdots + c_ne_n$$
for some $c_i \in \mathbb Z$. This doesn't mean that $\{e_1, ... , e_n\}$ or any subset of it has to be a basis. There exist finitely generately abelian groups which do not have any basis.
Example: $G = \mathbb Z/3\mathbb Z \times \mathbb Z/3\mathbb Z$ is finitely generated, by $e_1 = (1,0)$ and $e_2 = (0,1)$. But no subset of $\{e_1, e_2\}$ is a basis for $G$.
Best Answer
Proof with no structure: let $G$ be a f.g. divisible abelian group. Assuming by contradiction that $G\neq 0$, it has a maximal (proper) subgroup $H$; then $G/H$ is a simple abelian group, hence cyclic of prime order, hence is not divisible, contradiction since being divisible passes to quotients.
The same shows that a nonzero f.g. module over any commutative ring $A$ that is not a field, is never divisible (where divisible means that $m\mapsto am$ is surjective for every nonzero $a\in A$).