Hartshorne Exercise III 6.2 (a) Let $X=\mathbb{P}^1_k$, with $k$ an infinite field. Show there does not exist a projective object $\mathcal{P}\to\mathcal{O}_X\to 0$. 
The author suggests to consider surjections of the form $j_!(O_{X}|_{V})\to k(x)\to 0$, where $x$ is a closed point, $V$ an open neighbourhood of $x$.
We have maps $\mathcal{P}\to j_!(O_{X}|_{V})$ by lifting property, but how to use the conditions $\mathbb{P}^1_k$, with $k$ being infinite and $x$ being closed?
 A: This is an adaptation of Jon Woolf's answer at https://mathoverflow.net/a/5401/15505. Also, I didn't require $k$ to be infinite, which makes me uneasy.
Suppose there exists a projective object $\mathcal{P} \in \mathfrak{Mod}(X)$ and a morphism $\varphi : \mathcal{P} \to \mathcal{O}_{X}$; we show that $\varphi$ is the zero morphism. It suffices to show that $\varphi(U) : \mathcal{P}(U) \to \mathcal{O}_{X}(U)$ is the zero map for all nonempty open subsets $U$. Let $U$ be a nonempty open subset of $X$ with open immersion $\alpha_{U} : U \to X$; then $U$ is the complement of a finite number of closed points. Since closed points are dense in $\mathbb{P}_{k}^{1}$, we can choose two distinct closed points $x_{1},x_{2}$ of $U$; define $V_{1} = X \setminus \{x_{1}\}$ and $V_{2} = X \setminus \{x_{2}\}$; then $X = V_{1} \cup V_{2}$ and $U \not\subseteq V_{1}$ and $U \not\subseteq V_{2}$. There is a canonical morphism of presheaves $\alpha_{V_{1},!}^{\operatorname{pre}}\mathcal{O}_{V_{1}} \oplus \alpha_{V_{2},!}^{\operatorname{pre}}\mathcal{O}_{V_{2}} \to \mathcal{O}_{X}$ which is surjective on the stalks; this induces a surjective morphism $\xi : \alpha_{V_{1},!}\mathcal{O}_{V_{1}} \oplus \alpha_{V_{2},!}\mathcal{O}_{V_{2}} \to \mathcal{O}_{X}$ of sheaves; here we have $\alpha_{V_{1},!}^{\operatorname{pre}}\mathcal{O}_{V_{1}} = \alpha_{V_{1},!}\mathcal{O}_{V_{1}}$ and $\alpha_{V_{2},!}^{\operatorname{pre}}\mathcal{O}_{V_{2}} = \alpha_{V_{2},!}\mathcal{O}_{V_{2}}$ by https://math.stackexchange.com/a/1066689/7719 since $X$ is irreducible and $V_{1}$ is an integral scheme. By projectivity of $\mathcal{P}$, we have that $\varphi$ factors through $\xi$, via some morphism $\gamma : \mathcal{P} \to \alpha_{V_{1},!}\mathcal{O}_{V_{1}} \oplus \alpha_{V_{2},!}\mathcal{O}_{V_{2}}$ such that $\varphi = \xi \gamma$. Then we have $(\alpha_{V_{1},!}\mathcal{O}_{V_{1}} \oplus \alpha_{V_{2},!}\mathcal{O}_{V_{2}})(U) = (\alpha_{V_{1},!}^{\operatorname{pre}}\mathcal{O}_{V_{1}})(U) \oplus (\alpha_{V_{2},!}^{\operatorname{pre}}\mathcal{O}_{V_{2}})(U) = 0$, which implies that $\varphi(U) = 0$. 
