Let $\mathfrak m$ be the maximal ideal of $R$. We have to show that $$ \Mor_S(\Spec(R), X) \longrightarrow \lim \Mor_S(\Spec(R/\mathfrak m^n), X) $$ is bijective for $R$ as above. \medskip\noindent Injectivity: Let $x, x' : \Spec(R) \to X$ be two morphisms mapping to the same element in the right hand side. Consider the fibre product $$ T = \Spec(R) \times_{(x, x'), X \times_S X, \Delta} X $$ Then $T$ is a scheme and $T \to \Spec(R)$ is locally of finite type, monomorphism, separated, and locally quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-properties-diagonal}. In particular $T$ is locally Noetherian, see Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}. Let $t \in T$ be the unique point mapping to the closed point of $\Spec(R)$ which exists as $x$ and $x'$ agree over $R/\mathfrak m$. Then $R \to \mathcal{O}_{T, t}$ is a local ring map of Noetherian rings such that $R/\mathfrak m^n \to \mathcal{O}_{T, t}/\mathfrak m^n\mathcal{O}_{T, t}$ is an isomorphism for all $n$ (because $x$ and $x'$ agree over $\Spec(R/\mathfrak m^n)$ for all $n$). Since $\mathcal{O}_{T, t}$ maps injectively into its completion (see Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}) we conclude that $R = \mathcal{O}_{T, t}$. Hence $x$ and $x'$ agree over $R$. \medskip\noindent Surjectivity: Let $(x_n)$ be an element of the right hand side. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$.  Denote $x_0 : \Spec(k) \to X$ the morphism induced on the residue field $k = R/\mathfrak m$. The morphism of schemes $U \times_{X, x_0} \Spec(k) \to \Spec(k)$ is surjective \'etale. Thus $U \times_{X, x_0} \Spec(k)$ is a nonempty disjoint union of spectra of finite separable field extensions of $k$, see Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}. Hence we can find a finite separable field extension $k \subset k'$ and a $k'$-point $u_0 : \Spec(k') \to U$ such that $$ \xymatrix{ \Spec(k') \ar[d] \ar[r]_-{u_0} & U \ar[d] \\ \Spec(k) \ar[r]^-{x_0} & X } $$ commutes. Let $R \subset R'$ be the finite \'etale extension of Noetherian complete local rings which induces $k \subset k'$ on residue fields (see Algebra, Lemmas \ref{algebra-lemma-henselian-cat-finite-etale} and \ref{algebra-lemma-complete-henselian}). Denote $x'_n$ the restriction of $x_n$ to $\Spec(R'/\mathfrak m^nR')$. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-etale-formally-etale} we can find an element $(u'_n) \in \lim \Mor_S(\Spec(R'/\mathfrak m^nR'), U)$ mapping to $(x'_n)$. By Lemma \ref{lemma-map-into-scheme} the family $(u'_n)$ comes from a unique morphism $u' : \Spec(R') \to U$. Denote $x' : \Spec(R') \to X$ the composition. Note that $R' \otimes_R R'$ is a finite product of spectra of Noetherian complete local rings to which our current discussion applies. Hence the diagram $$ \xymatrix{ \Spec(R' \otimes_R R') \ar[r] \ar[d] & \Spec(R') \ar[d]^{x'} \\ \Spec(R') \ar[r]^{x'} & X } $$ is commutative by the injectivity shown above and the fact that $x'_n$ is the restriction of $x_n$ which is defined over $R/\mathfrak m^n$. Since $\{\Spec(R') \to \Spec(R)\}$ is an fppf covering we conclude that $x'$ descends to a morphism $x : \Spec(R) \to X$. We omit the proof that $x_n$ is the restriction of $x$ to $\Spec(R/\mathfrak m^n)$.