Proof.
We write out the construction for the case that $R$ is Noetherian and $R\to S$ is of finite type. Without those assumptions, the proof is the same, except that we have to use some set (possibly infinite) of generators in each degree.
Start of the construction: Let $A(0) = R[x_1, \ldots , x_ n]$ be a (usual) polynomial ring and let $A(0) \to S$ be a surjection. As grading we take $A(0)_0 = A(0)$ and $A(0)_ d = 0$ for $d \not= 0$. Thus $\text{d} = 0$ and $\gamma _ n$, $n > 0$, is zero as well.
Choose generators $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$ for the kernel of the given map $A(0) = R[x_1, \ldots , x_ n] \to S$. We apply Example 23.6.2 $m$ times to get
\[ A(1) = A(0)\langle T_1, \ldots , T_ m\rangle \]
with $\deg (T_ i) = 1$ as a graded divided power polynomial algebra. We set $\text{d}(T_ i) = f_ i$. Since $A(1)$ is a divided power polynomial algebra over $A(0)$ and since $\text{d}(f_ i) = 0$ this extends uniquely to a differential on $A(1)$ by Lemma 23.6.8.
Induction hypothesis: Assume we are given factorizations
\[ R \to A(0) \to A(1) \to \ldots \to A(m) \to S \]
where $A(0)$ and $A(1)$ are as above and each $R \to A(m') \to S$ for $2 \leq m' \leq m$ satisfies properties (1) and (4) of the statement of the lemma and (2) replaced by the condition that $H_ i(A(m')) \to H_ i(S)$ is an isomorphism for $m' > i \geq 0$. The base case is $m = 1$.
Induction step: Assume we have $R \to A(m) \to S$ as in the induction hypothesis. Consider the group $H_ m(A(m))$. This is a module over $H_0(A(m)) = S$. In fact, it is a subquotient of $A(m)_ m$ which is a finite type module over $A(m)_0 = R[x_1, \ldots , x_ n]$. Thus we can pick finitely many elements
\[ e_1, \ldots , e_ t \in \mathop{\mathrm{Ker}}(\text{d} : A(m)_ m \to A(m)_{m - 1}) \]
which map to generators of this module. Applying Example 23.6.2 or 23.6.3 $t$ times we get
\[ A(m + 1) = A(m)\langle T_1, \ldots , T_ t\rangle \]
with $\deg (T_ i) = m + 1$ as a graded divided power algebra. We set $\text{d}(T_ i) = e_ i$. Since $A(m+1)$ is a divided power polynomial algebra over $A(m)$ and since $\text{d}(e_ i) = 0$ this extends uniquely to a differential on $A(m + 1)$ compatible with the divided power structure. Since we've added only material in degree $m + 1$ and higher we see that $H_ i(A(m + 1)) = H_ i(A(m))$ for $i < m$. Moreover, it is clear that $H_ m(A(m + 1)) = 0$ by construction.
To finish the proof we observe that we have shown there exists a sequence of maps
\[ R \to A(0) \to A(1) \to \ldots \to A(m) \to A(m + 1) \to \ldots \to S \]
and to finish the proof we set $A = \mathop{\mathrm{colim}}\nolimits A(m)$.
$\square$
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