Show that if $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module.

Question

As the question says, I want to show that

If $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module.

Now, here is my reasoning:

Let $F$ be a free $\mathbb{Z}$-module. Then, for any basis $$A \subset F$$ of $F$ we have that $$F \cong \bigoplus_{a \in A} \mathbb{Z}.$$

Now, I think we can say that $$S = \bigcap_{n = 1}^{\infty} nF = 0$$ since $$S = \bigcap_{i = 1}^{\infty} \cong \bigcap_{n = 1}^{\infty} n \Big(\bigoplus_{a \in A}\mathbb{Z}\Big).$$

Now, it is clear this must be $0$, since, take any element $h$ supposedly contained in $S$;

Then $$h = (\ldots,z_{\alpha},\ldots)$$ with only finitely many nonzero-entries. Prime-factorize each $z_{\alpha}$. Since there are infinitely many primes, and the union of the primes in each entry will be a union of finite sets (just picking out the prime factors, disregarding the possible powers), hence finite, we can choose a prime $p$ such that $p$ is not in the prime-factorization of any entry of $h$, hence $$h \not \in p\bigoplus_{a \in A} \mathbb{Z} \implies h \not \in S \quad(\text{contradiction!}).$$

Now, I think there are two ways to show that if $Q$ was projective, we would get a contradiction. I'll just write what I think is the easier one; note that if $R$ (here $\mathbb{Z}$) is a P.I.D. then free is equivalent to projective, so that if $Q$ was a projective $\mathbb{Z}$-module, it would have to be free, hence we would get $$Q \cong \bigoplus_{b \in \beta} \mathbb{Z}$$ for some basis $\beta \subset Q$.

Now, since $Q$ is a divisible $\mathbb{Z}$-module, we have $$nQ = Q \quad(\forall n \in \mathbb{Z},n \neq 0)$$ it follows that $$\bigcap_{n = 1}^{\infty} nQ = Q$$ but $$\bigcap_{n = 1} nQ \cong \bigcap_{n = 1}^{\infty} n\Big(\bigoplus_{b \in \beta}\mathbb{Z}\Big) = 0 \quad(\text{contradiction!})$$

Is this proof correct?