Observe that $X$ is Cohen-Macaulay, see Algebra, Lemma \ref{algebra-lemma-lci-CM}. Thus $X = X' \amalg X''$ with $\dim(X') = 0$ and $X''$ equidimensional of dimension $1$, see Morphisms, Lemma \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension}. Since $X'$ is finite over $k$ (Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}) we can find $Y' \to \Spec(k[[t]])$ with special fibre $X'$ and generic fibre smooth by Lemma \ref{lemma-smoothing-artinian-lci}. Thus it suffices to prove the lemma for $X''$. After replacing $X$ by $X''$ we have $X$ is Cohen-Macaulay and equidimensional of dimension $1$. \medskip\noindent We are going to use deformation theory for the situation $\Lambda = k \to k$. Let $p_1, \ldots, p_r \in X$ be the closed singular points of $X$, i.e., the points where $X \to \Spec(k)$ isn't smooth. For each $i$ we pick an integer $n_i$ and a flat, essentially of finite type ring map $$ k[[t]] \longrightarrow B_i $$ with $B_i/tB_i \cong \mathcal{O}_{X, p_i}$ such that $t^{n_i}$ is in the $1$st Fitting ideal of $\Omega_{B_i/k[[t]]}$. This is possible by Lemma \ref{lemma-smoothing-at-lci-point}. Observe that the system $(B_i/t^nB_i)$ defines a formal object of $\Deformationcategory_{\mathcal{O}_{X, p_i}}$ over $k[[t]]$. By Lemma \ref{lemma-curve-isolated} the map $$ \Deformationcategory_X \longrightarrow \prod\nolimits_{i = 1, \ldots, r} \Deformationcategory_{\mathcal{O}_{X, p_i}} $$ is a smooth map between deformation categories. Hence by Formal Deformation Theory, Lemma \ref{formal-defos-lemma-smooth-morphism-essentially-surjective} there exists a formal object $(X_n)$ in $\Deformationcategory_X$ mapping to the formal object $\prod_i (B_i/t^n)$ by the arrow above. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-formal-algebraic-space-proper-reldim-1} there exists a projective scheme $Y$ over $k[[t]]$ and compatible isomorphisms $Y \times_{\Spec(k[[t]])} \Spec(k[t]/(t^n)) \cong X_n$. By More on Morphisms, Lemma \ref{more-morphisms-lemma-check-flatness-on-infinitesimal-nbhds} we see that $Y \to \Spec(k[[t]])$ is flat. Since $X$ is Cohen-Macaulay and equidimensional of dimension $1$ we may apply Lemma \ref{lemma-criterion-smoothing} to check $Y$ has smooth generic fibre\footnote{Warning: in general it is {\bf not} true that the local ring of $Y$ at the point $p_i$ is isomorphic to $B_i$. We only know that this is true after dividing by $t^n$ on both sides!}. Choose $n$ strictly larger than the maximum of the integers $n_i$ found above. It we can show $t^{n - 1}$ is in the first Fitting ideal of $\Omega_{X_n/S_n}$ with $S_n = \Spec(k[t]/(t^n))$, then the proof is done. To do this it suffices to prove this is true in each of the local rings of $X_n$ at closed points $p$. However, if $p$ corresponds to a smooth point for $X \to \Spec(k)$, then $\Omega_{X_n/S_n, p}$ is free of rank $1$ and the first Fitting ideal is equal to the local ring. If $p = p_i$ for some $i$, then $$ \Omega_{X_n/S_n, p_i} = \Omega_{(B_i/t^nB_i)/(k[t]/(t^n))} = \Omega_{B_i/k[[t]]}/t^n\Omega_{B_i/k[[t]]} $$ Since taking Fitting ideals commutes with base change (with already used this but in this algebraic setting it follows from More on Algebra, Lemma \ref{more-algebra-lemma-fitting-ideal-basics}), and since $n - 1 \geq n_i$ we see that $t^{n - 1}$ is in the Fitting ideal of this module over $B_i/t^nB_i$ as desired.