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Here is another proof:

Let $f:X\to Y$ be finite, $L$ an ample line bundle on $Y$, and let $F$ be a coherent sheaf on $X$. We may assume that $X$ and $Y$ are irreducible. Assume that $X$ is generically reduced.

We want to prove that $F\otimes (f^*L)^m$ is generated by global sections for $m\gg 0$.

Let $x\in X$ and $s\in F_x$. By noetherian induction we may assume that $x\in X$ is general and hence $X$ is locally irreducible and we may assume that $s$ is not torsion. Choose a (small enough irreducible) neighborhood $x\in U\subseteq X$ such that we may assume that $s\in (F\otimes L^m)(U)$. Let $Z=X\setminus U$. Since $f$ is proper, $f(Z)\subsetneq Y$ is closed. Let $V=Y\setminus f(Z)$, a non-empty open set. Then $f^{-1}V\subseteq U$ is a non-empty and hence dense open set in $X$.

Consider an arbitrary $y\in V$ and the image of $s\in (F\otimes L^m)(U)\to(f_*F\otimes L^m)(V)\to (f_*F)_y$ via restriction. By assumption $L$ is ample, so $f_*F\otimes L^m$ is generated by global sections for $m\gg 0$, so there exists a $\sigma\in H^0(Y, f_*F\otimes L^m)$ such that its image in $(f_*F)_y$ is the above element.

Observe that $\sigma\in H^0(X, F\otimes (f^*L)^m)$ and the above means that its restriction to $U$ agrees with $s$ on a non-empty open subset. But then it has to agree everywhere on $U$, in particular its germ at $x$ has to be the same as the original $s$.

1

Here is another proof:

Let $f:X\to Y$ be finite, $L$ an ample line bundle on $Y$, and let $F$ be a coherent sheaf on $X$. We may assume that $X$ and $Y$ are irreducible.

We want to prove that $F\otimes (f^*L)^m$ is generated by global sections for $m\gg 0$.

Let $x\in X$ and $s\in F_x$. Choose a (small enough) neighborhood $x\in U\subseteq X$ such that we may assume that $s\in (F\otimes L^m)(U)$. Let $Z=X\setminus U$. Since $f$ is proper, $f(Z)\subsetneq Y$ is closed. Let $V=Y\setminus f(Z)$, a non-empty open set. Then $f^{-1}V\subseteq U$ is a non-empty and hence dense open set in $X$.

Consider an arbitrary $y\in V$ and the image of $s\in (F\otimes L^m)(U)\to(f_*F\otimes L^m)(V)\to (f_*F)_y$ via restriction. By assumption $L$ is ample, so $f_*F\otimes L^m$ is generated by global sections for $m\gg 0$, so there exists a $\sigma\in H^0(Y, f_*F\otimes L^m)$ such that its image in $(f_*F)_y$ is the above element.

Observe that $\sigma\in H^0(X, F\otimes (f^*L)^m)$ and the above means that its restriction to $U$ agrees with $s$ on a non-empty open subset. But then it has to agree everywhere on $U$, in particular its germ at $x$ has to be the same as the original $s$.