Choose an affine open $\Spec(\Lambda) \subset S$ such that $k$ is a finite $\Lambda$-algebra, see Morphisms, Lemma \ref{morphisms-lemma-point-finite-type}. We may and do replace $S$ by $\Spec(\Lambda)$. \medskip\noindent We may write $R$ as a directed colimit $R = \colim C_j$ where each $C_j$ is a finite type $\Lambda$-algebra (see Algebra, Lemma \ref{algebra-lemma-ring-colimit-fp}). By assumption (b) the object $x$ is isomorphic to the restriction of an object over one of the $C_j$. Hence we may choose a finite type $\Lambda$-algebra $C$, a $\Lambda$-algebra map $C \to R$, and an object $x_C$ of $\mathcal{X}$ over $\Spec(C)$ such that $x = x_C|_{\Spec(R)}$. The choice of $C$ is a bookkeeping device and could be avoided. For later use, let us write $C = \Lambda[y_1, \ldots, y_u]/(f_1, \ldots, f_v)$ and we denote $\overline{a}_i \in R$ the image of $y_i$ under the map $C \to R$. Set $\mathfrak m_C = C \cap \mathfrak m_R$. \medskip\noindent Choose a $\Lambda$-algebra surjection $\Lambda[x_1, \ldots, x_s] \to k$ and denote $\mathfrak m'$ the kernel. By the universal property of polynomial rings we may lift this to a $\Lambda$-algebra map $\Lambda[x_1, \ldots, x_s] \to R$. We add some variables (i.e., we increase $s$ a bit) mapping to generators of $\mathfrak m_R$. Having done this we see that $\Lambda[x_1, \ldots, x_s] \to R/\mathfrak m_R^2$ is surjective. Then we see that \begin{equation} \label{equation-surjection} P = \Lambda[x_1, \ldots, x_s]_{\mathfrak m'}^\wedge \longrightarrow R \end{equation} is a surjective map of Noetherian complete local rings, see for example Formal Deformation Theory, Lemma \ref{formal-defos-lemma-surjective-cotangent-space}. \medskip\noindent Choose lifts $a_i \in P$ of $\overline{a}_i$ we found above. Choose generators $b_1, \ldots, b_r \in P$ for the kernel of (\ref{equation-surjection}). Choose $c_{ji} \in P$ such that $$ f_j(a_1, \ldots, a_u) = \sum c_{ji} b_i $$ in $P$ which is possible by the choices made so far. Choose generators $$ k_1, \ldots, k_t \in \Ker(P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P) $$ and write $k_i = (k_{i1}, \ldots, k_{ir})$ and $K = (k_{ij})$ so that $$ P^{\oplus t} \xrightarrow{K} P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P \to R \to 0 $$ is an exact sequence of $P$-modules. In particular we have $\sum k_{ij} b_j = 0$. After possibly increasing $N$ we may assume $N - 1$ works in the Artin-Rees lemma for the first two maps of this exact sequence (see More on Algebra, Section \ref{more-algebra-section-artin-rees} for terminology). \medskip\noindent By assumption $\mathcal{O}_{S, s} = \Lambda_{\Lambda \cap \mathfrak m'}$ is a G-ring. Hence by More on Algebra, Proposition \ref{more-algebra-proposition-finite-type-over-G-ring} the ring $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$ is a $G$-ring. Hence by Smoothing Ring Maps, Theorem \ref{smoothing-theorem-approximation-property-variant} there exist an \'etale ring map $$ \Lambda[x_1, \ldots, x_s]_{\mathfrak m'} \to B, $$ a maximal ideal $\mathfrak m_B$ of $B$ lying over $\mathfrak m'$, and elements $a'_i, b'_i, c'_{ij}, k'_{ij} \in B'$ such that \begin{enumerate} \item $\kappa(\mathfrak m') = \kappa(\mathfrak m_B)$ which implies that $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'} \subset B_{\mathfrak m_B} \subset P$ and $P$ is identified with the completion of $B$ at $\mathfrak m_B$, see remark preceding Smoothing Ring Maps, Theorem \ref{smoothing-theorem-approximation-property-variant}, \item $a_i - a'_i, b_i - b'_i, c_{ij} - c'_{ij}, k_{ij} - k'_{ij} \in (\mathfrak m')^N P$, and \item $f_j(a'_1, \ldots, a'_u) = \sum c'_{ji} b'_i$ and $\sum k'_{ij}b'_j = 0$. \end{enumerate} Set $A = B/(b'_1, \ldots, b'_r)$ and denote $\mathfrak m_A$ the image of $\mathfrak m_B$ in $A$. (Note that $A$ is essentially of finite type over $\Lambda$; at the end of the proof we will show how to obtain an $A$ which is of finite type over $\Lambda$.) There is a ring map $C \to A$ sending $y_i \mapsto a'_i$ because the $a'_i$ satisfy the desired equations modulo $(b'_1, \ldots, b'_r)$. Note that $A/\mathfrak m_A^N = R/\mathfrak m_R^N$ as quotients of $P = B^\wedge$ by property (2) above. Set $x_A = x_C|_{\Spec(A)}$. Since the maps $$ C \to A \to A/\mathfrak m_A^N \cong R/\mathfrak m_R^N \quad\text{and}\quad C \to R \to R/\mathfrak m_R^N $$ are equal we see that $x_A$ and $x$ agree modulo $\mathfrak m_R^N$ via the isomorphism $A/\mathfrak m_A^N = R/\mathfrak m_R^N$. At this point we have shown properties (1) -- (5) of the statement of the lemma. To see (6) note that $$ P^{\oplus t} \xrightarrow{K} P^{\oplus r} \xrightarrow{(b_1, \ldots, b_r)} P \quad\text{and}\quad P^{\oplus t} \xrightarrow{K'} P^{\oplus r} \xrightarrow{(b'_1, \ldots, b'_r)} P $$ are two complexes of $P$-modules which are congruent modulo $(\mathfrak m')^N$ with the first one being exact. By our choice of $N$ above we see from More on Algebra, Lemma \ref{more-algebra-lemma-approximate-complex-graded} that $R = P/(b_1, \ldots, b_r)$ and $P/(b'_1, \ldots, b'_r) = B^\wedge/(b'_1, \ldots, b'_r) = A^\wedge$ have isomorphic associated graded algebras, which is what we wanted to show. \medskip\noindent This last paragraph of the proof serves to clean up the issue that $A$ is essentially of finite type over $S$ and not yet of finite type. The construction above gives $A = B/(b'_1, \ldots, b'_r)$ and $\mathfrak m_A \subset A$ with $B$ \'etale over $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$. Hence $A$ is of finite type over the Noetherian ring $\Lambda[x_1, \ldots, x_s]_{\mathfrak m'}$. Thus we can write $A = (A_0)_{\mathfrak m'}$ for some finite type $\Lambda[x_1, \ldots, x_n]$ algebra $A_0$. Then $A = \colim (A_0)_f$ where $f \in \Lambda[x_1, \ldots, x_n] \setminus \mathfrak m'$, see Algebra, Lemma \ref{algebra-lemma-localization-colimit}. Because $p : \mathcal{X} \to (\Sch/S)_{fppf}$ is limit preserving on objects, we see that $x_A$ comes from some object $x_{(A_0)_f}$ over $\Spec((A_0)_f)$ for an $f$ as above. After replacing $A$ by $(A_0)_f$ and $x_A$ by $x_{(A_0)_f}$ and $\mathfrak m_A$ by $(A_0)_f \cap \mathfrak m_A$ the proof is finished.