[proofplan] A holomorphic vector bundle is the same thing as a locally free coherent analytic sheaf. GAGA says every coherent analytic sheaf on a projective variety is the analytification of a unique coherent algebraic sheaf. It remains only to check that local freeness descends from the analytic sheaf to the algebraic one. [/proofplan] [step:Translate the bundle into a coherent analytic sheaf] Let $E\to X^{\mathrm{an}}$ be a holomorphic vector bundle. Its sheaf of holomorphic sections $\mathcal{O}(E)$ is locally isomorphic to $\mathcal{O}_{X^{\mathrm{an}}}^r$, so it is a locally free coherent analytic sheaf. [/step] [step:Apply GAGA] By Serre's GAGA theorem, there exists a coherent algebraic sheaf $\mathcal{F}$ on $X$, unique up to unique isomorphism, such that \begin{align*} \mathcal{F}^{\mathrm{an}}\cong \mathcal{O}(E). \end{align*} The uniqueness part of GAGA also says that any two algebraic sheaves with this analytification are algebraically isomorphic. [/step] [step:Show the algebraic sheaf is locally free] Local freeness can be checked on completed local rings, or equivalently after passing to the associated analytic local rings. Since $\mathcal{F}^{\mathrm{an}}$ is locally free of rank $r$, the stalk $\mathcal{F}_x\otimes_{\mathcal{O}_{X,x}}\mathcal{O}_{X^{\mathrm{an}},x}$ is free of rank $r$ for every $x$. Faithful flatness of analytification of local rings implies that $\mathcal{F}_x$ itself is free of rank $r$. Thus $\mathcal{F}$ is a locally free algebraic sheaf. [/step] [step:Return to vector bundles] Locally free algebraic sheaves of rank $r$ are equivalent to algebraic rank-$r$ vector bundles. The locally free sheaf $\mathcal{F}$ therefore defines an algebraic vector bundle whose analytification has sheaf of holomorphic sections $\mathcal{O}(E)$. Hence it analytifies to $E$, and uniqueness follows from the full faithfulness part of GAGA. [/step]