I've seen this statement on the internet but I could not find a proof. Actually this is true for any module I think. Can a proof be given as follows?
Let $M$ be an $R$-module. Take a generating set $X$ of $M$ over $R$. Then consider the free $R$-module $F$ over the set $X$. Hence $M \subset F$ and we are done. I hope this is not a nonsense idea.
What you get with your $X$ (a generating set of $M$) and your $F$ (free on $X$) is a surjective homomorphism $\pi:F\to M$, not an injection $M\to F$.
But if $M$ is projective, the surjection $\pi$ splits, that is there's an injective homomorphism $\iota:M\to F$ with $\pi\circ\iota=\text{id}_M$.