I want to determine whether the following statement is true:
Let $R$ be any ring, and $V$ an $R$-module that is a union $V=\bigcup_{n=1}^\infty V_n$ of submodules $V_1 \subseteq V_2 \subseteq \dots$. If each $V_i$ is projective, then so is $V$.
My attempt at a proof is to use Zorn's lemma as follows: let $g : M \twoheadrightarrow N$ be a surjective $R$-module homomorphism. Let $f: V \to N$ be a module homomorphism. Consider the set $S$ of pairs $(V_i,f_i)$, where $f_i : V_i \to M$ is such that $g\circ f_i = f\big|_{V_i}$. Define a partial order on $S$ by $(V_i,f_i)\leq (V_j,f_j)$ if and only if $V_i \subseteq V_j$ and $f_j\big|_{V_i} = f_i$. The union of any chain gives an upper bound, so by Zorn's lemma, there is a maximal element $(V_m,f_m)$ of $S$. The problem here is that I don't think $V_m = V$ necessarily, since we may not be able to extend this particular $f_m$ to a larger $V_i$.
A possible counterexample to the statement is as follows: Let $p_1,p_2,p_3,\dots$ be the distinct prime numbers of $\mathbb{Z}^{>0}$. Set $V_i = \left(\prod_{k=1}^i p_k^i\right)^{-1}\mathbb{Z}$. Then each $V_i$ is a free $\mathbb{Z}$-module, hence projective. We have $\mathbb{Q} = \bigcup_{i=1}^\infty V_i$, and $V_1 \subseteq V_2 \subseteq \dots$, but $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module.
Does this counterexample work? Is there an easier counterexample?
Your counterexample works fine.
A couple other counterexamples that might be a little simpler are $V_i=(i!)^{-1}\mathbb{Z}$ (here the union is $\mathbb{Q}$), or $V_i = n^{-i}\mathbb{Z}$ where $n\geq 2$ is fixed (here the union is the ring $\mathbb{Z}[n^{-1}]$).