A group is called finitely generated if it has a presentation with finite generators.
Edit: My original question was vacuous. Suppose that $G$ is a finitely generated group and $\{g_i\}_{i\in I}$ is a generating set of $G$. Then, my question is :
Can we choose finitely many $g_{i_k}$'s ($i_k\in I$) such that $\{g_{i_k}\}$ becomes a generating set of $G$?
You have a finite generating set $A$ and a potentially infinite generating set $B$.
Now every element of $A$ can be written as a finite expression in terms of elements of $B$. The elements of $B$ which appear in the (finitely many!) expressions we have found obviously also generate the group.
This is for an old, different version of the question. No. There is a presentation of $G=\mathbb Z$ which has one generator for each element of $G$, for example.
Indeed, for each $n\in\mathbb Z$ let $x_n$ be a letter, and consider the quotient of the free group on the set $\{x_n:n\in\mathbb Z\}$ modulo the normal subgroup generated by all elements of the form $x_nx_mx_{n+m}^{-1}$ with $n$, $m\in\mathbb Z$.
You can play all sort of silly tricks in this way. For example, the group $\mathbb Z$ can be presented as the quotient of the free group generated by letters $x_i$ with $i\in\mathbb R$ modulo the normal subgroup generated by all elements of the form $x_ix_j^{-1}$ with $i, j\in\mathbb R$.