Here a slightly different proof. Decompose $\hat{\mathbb{Z}}$ into product of the pro-l Sylows. Now a vector in this decomposition having a zero coordinate at the prime p must be divisible by any integer coprime to p.

Any other element is the sum of two such elements, hence you are done (or alternatively fix only one prime, apply the above, now you have a map from the p-adics to the integers and it must be 0, as the p-adics are l-divisible for any l coprime to p).

Here a slightly different proof. Decompose $\hat{\mathbb{Z}}$ into product of the pro-l Sylows. Now a vector in this decomposition having a zero coordinate at the prime p must be divisible by any power of p.

Any other element is the sum of two such elements, hence you are done (or alternatively fix only one prime, apply the above, now you have a map from the p-adics to the integers and it must be 0, as the p-adics are l-divisible for any l coprime to p).