**Proof.**
Let $Y_1, Y_2$ be finite étale over $X$ and let $\varphi : (Y_1)_ U \to (Y_2)_ U$ be a morphism over $U$. We have to show that $\varphi $ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma 58.10.1.

Let $x \in X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. Since every point of $X \setminus U$ is closed $V$ is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ uniquely to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. By Limits, Lemma 32.20.3 (this uses that $U$ is retrocompact in $X$) we find an open $U \subset U'_ x$ containing $x$ and an extension $\varphi '_ x : (Y_1)_{U'_ x} \to (Y_2)_{U'_ x}$ of $\varphi $. Note that given two points $x, x' \in X \setminus U$ the morphisms $\varphi '_ x$ and $\varphi '_{x'}$ agree over $U'_ x \cap U'_{x'}$ as $U$ is dense in that open (Lemma 58.10.1). Thus we can extend $\varphi $ to $\bigcup U'_ x = X$ as desired.
$\square$

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