I want to show that every generating set of $\mathbb{Q}$ (as $\mathbb{Z}$-module) contains infinite elements by showing that for every generating set $I$ and element $i \in I$ the set $I \setminus i$ generates the module as well.
Sadly I don't really know where to start.
Take $\frac i2\in \Bbb Q$, and write it out as a finite linear combination of elements of $I$ with integer coefficients. There might be a term with $i$ in there. Subtract it away. You now have a linear combination representing some $\frac{ki}2$, where $k$ is an odd integer. Multiply it by $2$, and you have a representation of $ki$ without using $i$. If $k = \pm1$ you're done.
Otherwise, consider $\frac{i}{k}$, and do exactly the same thing. Note that when you subtract away the term with $i$ from the linear combination, you end up with a linear combination representing some number of the form $\frac{mki + i}{k}$ for some integer $m$. Multiply by $k$, and you now have a representation of $mki + i$ without using $i$.
Take the representation of $ki$ from earlier, multiply that by $m$, and you get a representation of $mki$ without using $i$. So now we have a representation of $mki + i$, and a representation of $mki$, both without using $i$. If we subtract one from the other, we've got a representation of $i$ without using $i$, and we're done.