Suppose that $A$ is a Dedekind domain (and integral domain). I am trying to prove that if $M$ is a torsion-free $A$-module, then it is projective and vice versa.
Suppose that $M$ is projective. Then it is well-known that $M$ is a direct summand of a free module and hence it is a submodule of free module $F$. Now suppose $a \in A$ such that $a \neq 0$ and $m \in M$. Since $F$ is free, we can write $m = a_1 f_1 + ... + a_k f_k$. Suppose $am = 0$, so $a a_1 f_1 +...+aa_k f_k = 0$, and $a a_i = 0$, and we have $a_i = 0$, since $A$ is an integral domain. It follows that $m = 0$ and this is a contradiction, so $M$ is torsion-free.
I was trying to prove then that "torsion-free $\Rightarrow$ projective" by using a similar argument, however I got stuck. Could you please suggest, how to prove the other direction?
Finitely generated and torsion-free implies projective, but infinitely generated torsion-free modules need not be projective. (E.g. the fraction field of $A$ is not projective as an $A$-module; think about $\mathbb Q$ over $\mathbb Z$.)
This is why you are having trouble proving it; you have to use finitely generated in the argument. (More generally, over any commutative ring with $1$, f.p. and flat is equivalent to f.g. and projective.)