Q as an additive abelian group has no minimal generating set 
$\mathbb Q$ as an additive abelian group has no minimal generating set.

I have done this question according to the solution given here. First I took a minimal generating set $S$ of $\mathbb Q$ and an element $a\in S$. Then I showed the set $T=S\setminus \{a\}$ is a generating set for $\mathbb Q$.
A student asked me that following in this way we can eliminate infinitely many elements and reach a stage where we are left with a generating set for $\mathbb Q$ having finitely many elements and this will contradict the fact that $\mathbb Q$ as an additive abelian group is not finitely generated.
I know we just cannot reach to a finitely generating set in this way. But I don't know the correct explanation for why we cannot do this.
 A: This is to be read as a proof by contradiction, not of construction. You supply me with a set of generators that you think may be minimal, and I say "no, it's not, because this one is smaller" as I remove one of your generators. I can do that no matter which generating set you give me, which means that no single set of generators can be minimal.
If you do repeat the process, you can only remove a finite number of generators this way. That finite number may be taken to be as large as you want, but it must be finite. Besides, if you have an infinite set and remove an infinite subset, you have no guarantee that you end up with something finite.
A: After removing an infinite number of elements, you may be left with an infinite set, or a finite set, or even the empty set. In any case, there is no reason to expect that the set you are left with is a generating set. In general, the intersection of an infinite decreasing sequence of generating sets does not have to be a generating set. For example, you could start with the set $\mathbb Q$ of all rational numbers, and remove one number at a time, in such a way that after an infinite sequence of steps there is nothing left.
