Let $X$ be an integral, affine scheme. Are there any errors in the following argument that $\mathcal{O}_X\text{-Mod}$ does not have enough projectives?
Suppose $\mathcal{P}$ is projective and $\rho: \mathcal{P} \rightarrow \mathcal{O}_X$ is a surjection. Then there is a surjection $\rho_x:\mathcal{P} \rightarrow \mathcal{O}_{X,x}$ for any $x \in X$ (throughout, $\mathcal{O}_{X,x}$ is regarded as an $\mathcal{O}_X$-module supported at $x$).
Let $ 0 \neq \sigma \in \mathcal{O}_{X,x}$ for some $x \in X$. Then there is an open $V \subset X$ and $s \in \Gamma(V,\mathcal{P})$ such that $\rho_x(V)(s)=\sigma$. Choose an open neighborhood $U$ of $x$ such that $V \not\subset U$. Let $\mathcal{O}_U$ denote $i_!(\mathcal{O}_X|_U)$, where $i:U \rightarrow X$ is the inclusion. Then there is a surjection $r_x:O_U \rightarrow O_{X,x}$. Since $\mathcal{P}$ is projective, there is a lifting $\tau:\mathcal{P} \rightarrow \mathcal{O}_U$ such that $\rho_x = r_x \circ \tau$.
Since $\Gamma(V,\mathcal{O}_U)=0$, $\tau(V) = 0$. Thus $\sigma = \rho_x(V)(s) = r_x(V)(\tau(V)(s))=0$, a contradiction.
Your argument is almost correct except that there may not exist any open neighborhood $U$ of $x$ such that $V\not\subseteq U$. For instance, what if $V=\{x\}$?
Indeed, the result you are trying to prove is not true. For instance, if $X$ is Spec of a field, then $\mathcal{O}_X\text{-Mod}$ is the category of vector spaces over the field, in which every object is projective. Your argument does prove that the result is true if there exists a point $x\in X$ which has no smallest open neighborhood, since then your $U$ is guaranteed to exist.