Part 1:
Definition: Suppose G is an abelian group, and S is a subset of G generating the group H. For each prime p, the group H/pH is a vector space over the field Z/pZ of p elements. Consider the set S + pH = { s + pH : s in S } in the group H/pH. We say that S is linearly independent mod p if S + pH is a linearly independent set in the vector space H/pH over Z/pZ.
Proposition: Suppose G is an abelian group, and S is a subset of G that is linearly independent mod p for every prime p. Then S is a basis of a free abelian subgroup of G.
Proof: Suppose Σ as s = 0 for some integers as. Then reducing the equation mod p produces a linear dependence relation in the vector space H/pH. Hence each as ≡ 0 mod p, by the definition of linearly independent. In particular, as is an integer divisible by every prime p. The only such integer is 0, and as = 0 for all s in S. This means that S is "independent", as in, it is a basis of the free abelian subgroup it spans. □
As far as I can tell, you don't need to assume G is torsion-free (but, the subgroup generated by S is torsion-free), and you don't need to assume S is finite. Note that no subgroup can possibly by linearly independent mod p according to my definition. You want a subset S, not a subgroup, even though you really are quite interested in the subgroup H it generates.
Part 2:
This is sort of a restatement of the previous part.
Proposition: Let G be a free abelian group with basis S. Then no element of S is divisible by any prime number p.
Proof: Suppose x in S is divisible by p, that is, suppose there is some y in G such that py = x. Write y as a linear combination of S: y = Σ as s. Then x = py = Σ (pas) s. Since S is a basis, the coefficients are uniquely determined and so we get 1 = pax and 0 = pas for s ≠ x, s in S. These are just equations of integers, and canceling p one gets ax is the "integer" 1/p, a contradiction. □