Category O has enough projectives

By [F1] there is a projective $Q\in {\mathcal{O}}_C$ mapping onto $L(\mu )$; applying [F2] to that epimorphism, some indecomposable direct summand $P(\mu )$ of $Q$ is a projective cover of $L(\mu )$, unique up to isomorphism and indecomposable, and it lies in ${\mathcal{O}}_C$, hence in the linkage class of $\mu$.

Every object $X$ of $\mathcal{O}$ is a quotient of a finite direct sum of such projective covers. Induct on the length of a composition series $0=X_0\subseteq X_1\subseteq \cdots \subseteq X_n=X$. For $n=0$ the zero object is a quotient of the empty sum. For $n\ge 1$, assume $p:Q^{\mathrm{\prime }}\twoheadrightarrow X_{n-1}$ with $Q^{\mathrm{\prime }}$ a finite direct sum of projective covers, and let $L({\mu }_n)=X_n\mathrm{/}{X}_{n-1}$ with its projective cover ${\pi }_n:P({\mu }_n)\twoheadrightarrow L({\mu }_n)$ from step 1.1. Since $P({\mu }_n)$ is projective and $X_n\twoheadrightarrow L({\mu }_n)$ is an epimorphism, ${\pi }_n$ lifts to $\tilde{\pi }:P({\mu }_n)\to X_n$, and the sum morphism $Q^{\mathrm{\prime }}\oplus P({\mu }_n)\to X_n$ is an epimorphism: an element $x\in X_n$ differs from an element of the image of $\tilde{\pi }$ by an element of $X_{n-1}$, which lies in the image of $p$. Hence $X_n$ is a quotient of the finite direct sum $Q^{\mathrm{\prime }}\oplus P({\mu }_n)$ of projective covers.

By step 1.1 each simple $L(\mu )$ has an indecomposable projective cover lying in its linkage class, unique up to isomorphism, and by step 2.1 every object of $\mathcal{O}$ is a quotient of a finite direct sum of these projective covers; this is exactly the assertion that $\mathcal{O}$ has enough projectives.