Missing argument in the proof of the closedness of the graph of a morhpism between algebraic varieties

I am following a proof of the following statements: if $f:Xto Y$ is a morphism of varieties (say, quasi-projective), then the graph of $f$, i.e. the subset $Gamma_f:=operatorname{Im}(operatorname{id}_Xtimes Y)subseteq Xtimes Y$ is a closed subvariety of $Xtimes Y$, such that the co-restriction of $operatorname{id}_Xtimes Y$ yields an isomorphism. In the proof, the author takes ${V_j}$ an open affine cover of $Y$ and then ${U_{ij}}$ an open affine cover of each of the $f^{-1}(V_j)$. Then they proceed to show that if $f_{ij}:U_{ij}to V_j$ denotes the restricted and corestricted morphism, then $Gamma_{f_{ij}}=Gamma_fcap(U_{ij}times V_j)$. Up until this point I can follow the argument. However, it is then said that in order to prove that $Gamma_f$ is closed, it suffices to prove that each of the $Gamma_{f_{ij}}$ is closed; i.e. the problem is reduced to the affine case. However, to me this would oly prove that $Gamma_{f}$ is locally closed. Is there a quick argument that deduces closedness from locally closedness? Or is there a something substantially missing from the argument?

Disclaimer: I only yet know the classical approach to algebraic geometry, and not the modern, scheme theoretic approach