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First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.

BTW, just a silly but perhaps cute remark; one may apply this same argument to the map $\pi: *\to BG$, for $G$ a finite group, to obtain that every irreducible representation of $G$ is a subrepresentation of the regular representation (this is basically the same as the usual argument using the adjointness of induction and restriction).

First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.

BTW, just a silly but perhaps cute remark; one may apply this same argument to the map $\pi: *\to BG$, for $G$ a finite group, to obtain that every irreducible representation of $G$ is a subrepresentation of the regular representation (this is basically the same as the usual argument using the adjointness of induction and restriction).

First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.

First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.

BTW, just a silly but perhaps cute remark; one may apply this same argument to the map $\pi: *\to BG$, for $G$ a finite group, to obtain that every irreducible representation of $G$ is a subrepresentation of the regular representation (this is basically the same as the usual argument using the adjointness of induction and restriction).

Oops, what I originally wrote assumed U=X; didn't read the question properly. Here's a fix:

First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}|_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.

First, suppose $L$ is simple. $L$ is trivialied by some finite etale map $\tilde U\to U$ (this is what it means to have finite monodromy); then applying Grauert-Remmert (or SGA I, as mentioned in Georges Elencwajg's answer to this question), we may extend the map to a (possibly ramified) map $\pi: \tilde X\to X$. That is, $\pi^*L|_{\tilde U}=\underline{\mathbb{C}}^n$. Then $\operatorname{Hom}_U(L, \pi_*(\underline{\mathbb{C}}^n)|_{U})\simeq\operatorname{Hom}_{\tilde U}(\pi^*L|_{\tilde U}, \underline{\mathbb{C}}^n)$ is non-zero, so $L$ is a factor of $\pi_*\underline{\mathbb{C}}^n$ (using simplicity). Again, as $L$ is simple, it is in fact a factor of $\pi_*\underline{\mathbb{C}}$.

Now if $L$ is not simple, decompose it into simple factors $L=\oplus L_i$, and let $\pi: {\tilde X_i}\to X$ be covers trivializing the $L_i$. Setting $\tilde X=\bigsqcup {\tilde X_i}$ does the trick.