I have the following proof that a torsion-free module over a PID is flat.
Let $A$ be a PID, and let $N$ be a torsion-free $A$-module. If $N$ is finitely generated, then $N$ is free (by the classification of finitely generated modules over a PID), and it is easy to show that $N$ is flat. Otherwise, we can reduce to the finitely generated case as follows. Suppose toward a contradiction that $N$ is not flat. Then there exists an injective map $f : M' \to M$ of $A$-modules such that $$f \otimes \mathrm{Id}_N : M' \otimes N \to M \otimes N$$ is not injective. Let $\sum_{i = 1}^n x_i \otimes y_i$ be a nonzero element of $\ker(f \otimes \mathrm{Id}_N)$. Then $$ \sum_{i = 1}^n f(x_i) \otimes y_i = 0 $$ in $M \otimes N$, so we can find a finitely generated submodule $N'$ of $N$ containing $y_1, \dots, y_n$ such that $\sum_{i = 1}^n f(x_i) \otimes y_i = 0$ in $M \otimes N'$. Since $\sum_{i = 1}^n x_i \otimes y_i$ is nonzero in $M' \otimes N$, it is also nonzero in $M' \otimes N'$. Therefore, the map $$f \otimes \mathrm{Id}_{N'} : M' \otimes N' \to M \otimes N'$$ also is not injective, so $N'$ cannot be flat. However, $N'$ is both torsion-free and finitely generated, which contradicts our earlier observation.
Is this proof correct? I'm a bit skeptical since it seems a bit more elementary than the proofs presented here and here. The first answer uses the $\mathrm{Tor}$ functor, while the second answer uses the fact that $N$ is flat if and only if $- \otimes N$ preserves the injectivity of all inclusions $\mathfrak{a} \hookrightarrow A$ of ideals. (Edit: I just realized that I use the classification of finitely generated modules over a PID, so perhaps it's inaccurate to say that my proof is more elementary than the other two.)