Note that the flatness of $A \to \Lambda$ implies that $A \to \Lambda$ is injective, so the lemma really tells us that $\Lambda$ is a directed union of these types of subrings $A \subset \Lambda$. Let $n$ be the minimal integer such that $\mathfrak m^n = 0$. We will prove this lemma by induction on $n$. The case $n = 1$ is clear as a field extension is a union of finitely generated field extensions. \medskip\noindent Pick $\lambda_1, \ldots, \lambda_d \in \mathfrak m$ which generate $\mathfrak m$. As $K$ is formally smooth over $\mathbf{F}_p$ (see Algebra, Lemma \ref{algebra-lemma-formally-smooth-extensions-easy}) we can find a ring map $\sigma : K \to \Lambda$ which is a section of the quotient map $\Lambda \to K$. In general $\sigma$ is {\bf not} a $k$-algebra map. Given $\sigma$ we define $$ \Psi_\sigma : K[x_1, \ldots, x_d] \longrightarrow \Lambda $$ using $\sigma$ on elements of $K$ and mapping $x_i$ to $\lambda_i$. Claim: there exists a $\sigma : K \to \Lambda$ and a subfield $k \subset F \subset K$ finitely generated over $k$ such that the image of $k$ in $\Lambda$ is contained in $\Psi_\sigma(F[x_1, \ldots, x_d])$. \medskip\noindent We will prove the claim by induction on the least integer $n$ such that $\mathfrak m^n = 0$. It is clear for $n = 1$. If $n > 1$ set $I = \mathfrak m^{n - 1}$ and $\Lambda' = \Lambda/I$. By induction we may assume given $\sigma' : K \to \Lambda'$ and $k \subset F' \subset K$ finitely generated such that the image of $k \to \Lambda \to \Lambda'$ is contained in $A' = \Psi_{\sigma'}(F'[x_1, \ldots, x_d])$. Denote $\tau' : k \to A'$ the induced map. Choose a lift $\sigma : K \to \Lambda$ of $\sigma'$ (this is possible by the formal smoothness of $K/\mathbf{F}_p$ we mentioned above). For later reference we note that we can change $\sigma$ to $\sigma + D$ for some derivation $D : K \to I$. Set $A = F[x_1, \ldots, x_d]/(x_1, \ldots, x_d)^n$. Then $\Psi_\sigma$ induces a ring map $\Psi_\sigma : A \to \Lambda$. The composition with the quotient map $\Lambda \to \Lambda'$ induces a surjective map $A \to A'$ with nilpotent kernel. Choose a lift $\tau : k \to A$ of $\tau'$ (possible as $k/\mathbf{F}_p$ is formally smooth). Thus we obtain two maps $k \to \Lambda$, namely $\Psi_\sigma \circ \tau : k \to \Lambda$ and the given map $i : k \to \Lambda$. These maps agree modulo $I$, whence the difference is a derivation $\theta = i - \Psi_\sigma \circ \tau : k \to I$. Note that if we change $\sigma$ into $\sigma + D$ then we change $\theta$ into $\theta - D|_k$. \medskip\noindent Choose a set of elements $\{y_j\}_{j \in J}$ of $k$ whose differentials $\text{d}y_j$ form a basis of $\Omega_{k/\mathbf{F}_p}$. The Jacobi-Zariski sequence for $\mathbf{F}_p \subset k \subset K$ is $$ 0 \to H_1(L_{K/k}) \to \Omega_{k/\mathbf{F}_p} \otimes K \to \Omega_{K/\mathbf{F}_p} \to \Omega_{K/k} \to 0 $$ As $\dim H_1(L_{K/k}) < \infty$ we can find a finite subset $J_0 \subset J$ such that the image of the first map is contained in $\bigoplus_{j \in J_0} K\text{d}y_j$. Hence the elements $\text{d}y_j$, $j \in J \setminus J_0$ map to $K$-linearly independent elements of $\Omega_{K/\mathbf{F}_p}$. Therefore we can choose a $D : K \to I$ such that $\theta - D|_k = \xi \circ \text{d}$ where $\xi$ is a composition $$ \Omega_{k/\mathbf{F}_p} = \bigoplus\nolimits_{j \in J} k \text{d}y_j \longrightarrow \bigoplus\nolimits_{j \in J_0} k \text{d}y_j \longrightarrow I $$ Let $f_j = \xi(\text{d}y_j) \in I$ for $j \in J_0$. Change $\sigma$ into $\sigma + D$ as above. Then we see that $\theta(a) = \sum_{j \in J_0} a_j f_j$ for $a \in k$ where $\text{d}a = \sum a_j \text{d}y_j$ in $\Omega_{k/\mathbf{F}_p}$. Note that $I$ is generated by the monomials $\lambda^E = \lambda_1^{e_1} \ldots \lambda_d^{e_d}$ of total degree $|E| = \sum e_i = n - 1$ in $\lambda_1, \ldots, \lambda_d$. Write $f_j = \sum_E c_{j, E} \lambda^E$ with $c_{j, E} \in K$. Replace $F'$ by $F = F'(c_{j, E})$. Then the claim holds. \medskip\noindent Choose $\sigma$ and $F$ as in the claim. The kernel of $\Psi_\sigma$ is generated by finitely many polynomials $g_1, \ldots, g_t \in K[x_1, \ldots, x_d]$ and we may assume their coefficients are in $F$ after enlarging $F$ by adjoining finitely many elements. In this case it is clear that the map $A = F[x_1, \ldots, x_d]/(g_1, \ldots, g_t) \to K[x_1, \ldots, x_d]/(g_1, \ldots, g_t) = \Lambda$ is flat. By the claim $A$ is a $k$-subalgebra of $\Lambda$. It is clear that $\Lambda$ is the filtered colimit of these algebras, as $K$ is the filtered union of the subfields $F$. Finally, these algebras are essentially of finite type over $k$ by Algebra, Lemma \ref{algebra-lemma-essentially-of-finite-type-into-artinian-local}.