Proof.
Denote $\mathcal{C}$ the category introduced in the lemma. Observe that for $\varphi : U \to X$ in $\mathcal{C}$ the morphism $\varphi $ is representable (by schemes) and affine, see Lemma 87.19.7. Recall that $X_{affine, {\acute{e}tale}} = X_{red, affine, {\acute{e}tale}}$. Hence we can define a functor
\[ \mathcal{C} \longrightarrow X_{affine, {\acute{e}tale}},\quad (U \to X) \longmapsto U \times _ X X_{red} \]
because $U \times _ X X_{red}$ is an affine scheme.
To finish the proof we will construct a quasi-inverse. Namely, write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. For each $\lambda $ we have $X_{red} \subset X_\lambda $ is a thickening. Thus for every $\lambda $ we have an equivalence
\[ X_{red, affine, {\acute{e}tale}} = X_{\lambda , affine, {\acute{e}tale}} \]
for example by More on Algebra, Lemma 15.11.2. Hence if $U_{red} \to X_{red}$ is an étale morphism with $U_{red}$ affine, then we obtain a system of étale morphisms $U_\lambda \to X_\lambda $ of affine schemes compatible with the transition morphisms in the system defining $X$. Hence we can take
\[ U = \mathop{\mathrm{colim}}\nolimits U_\lambda \]
as our affine formal algebraic space over $X$. The construction gives that $U \times _ X X_\lambda = U_\lambda $. This shows that $U \to X$ is representable and étale. We omit the verification that the constructions are mutually inverse to each other.
$\square$
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