I would like to show that if $R$ is a ring with $1$, and $e$ is an idempotent in $R$, then $Re$ is a projective module.
My idea is to show that $R = Re \oplus R(1-e)$. It's clear to me that $R = Re + R(1-e)$, but I am having trouble showing that $Re \cap R(1-e) = \{0\}$. I tried that if $x \in Re \cap R(1-e)$, then $x = re = r'(1-e)$ for some $r,r' \in R$. Then $re = re^2 = r'(1-e)e = r'(e - e^2) = 0$, but how can I show $r$ is $0$ if $R$ need not necessarily be a domain?
As pointed out in the comments, your approach is very good. Here is another, which you might like.
There is an obvious surjective $R$-module homomorphism $\varphi \colon R \to Re$ given by $r \mapsto re$, so we obtain an exact sequence of $R$-modules
$$0 \to \ker(\varphi) \to R \xrightarrow{\varphi} Re \to 0$$
To show that $Re$ is a direct summand of $R$, it suffices to show that this sequence is split. The inclusion $i \colon Re \to R$ is a splitting, since $\varphi(i(re)) = re\cdot e = re^{2} = re$, so $R \cong \ker(\varphi) \oplus Re$.