Prove that $\mathbb{C}$ is not a projective $\mathbb{Z}$-module.
Suppose on the contrary that $\mathbb{C}$ is a projective $\mathbb{Z}$-module. Then we have that $\mathbb{C}$ is a submodule of $\mathbb{Z}^{(I)}$. So, the inclusion would send one in $\mathbb{C}$ to one in $\mathbb{Z}^{(I)}$. Then, $i(n/n)=n \hspace{0.02 in}i(1/n)$ implies that $(1/n)^{(I)}$ is an element of $\mathbb{Z}^{(I)}$, a contradiction. Is this proof correct?
The proof has the right idea but is not quite correct: the same logic shows that $\Bbb Z$ is not a $\Bbb Z$-module since $n=n(1/n)$ implies $1/n\in\Bbb Z$.
What you are getting at is divisibility: an $R$-module $V$ is divisible if $rV=V$ for all $r$ which are not zero divisors.
Over $\Bbb Z$, any free module is not divisible (see here and here). If $\Bbb C$ is free, this means that there exists $n\in\Bbb Z$ and $z\in\Bbb C$ for which $nz\notin\Bbb C$. Of course this is impossible since $\Bbb Z$ is a subring of $\Bbb C$ already. In other words, $\Bbb C$ is clearly divisible, hence not a free $\Bbb Z$-module.