Let $R$ be a commutative ring. Prove that $R/I$ is free if and only if $I=0.$
Is my following proof OK?
Assume $R/I$ is a free $R$-module. Then $\mathrm{Ann}(R/I)=0.$ That means $\{x \in R|x(r+I)=xr+I=0+I\},$ for all $r \in R.$ So $xr \in I$ for all $r \in R.$ So $x \in I.$ But $xr=0$ for all $r \in R.$ So $x=0.$ Since $x$ was arbitrary, $I=0.$
Assume $I=0.$ Then $\mathrm{Ann}(R/I)=\mathrm{Ann}(R)=0.$ So the zero element itself annihilates $R.$ Since the empty set is trivially a basis for $R/I, R/I$ is free.
The idea for the first implication is just fine: a free module is always faithful, but $R/I$ has annihilator $I$, so if it is free, $I=\{0\}$. Good job!
This isn't true, and doesn't follow any of the reasoning about the annihilator that you started with. The empty set is never a basis for a nonzero $R$ module. The empty set can only "span" $\{0\}$ in the sense that it is the smallest submodule containing the empty set.
Worrying about annihilators in that direction is off track anyhow. Just note that when $I=\{0\}$, $R/I\cong R$ which is free, so $R/I$ is free as well.