Write
$$
C=A_0A_1\cdots A_r.
$$
Under the improved bounds allowed in the question, $r=O(\log^{O(1)}(2K))$ and
$$
|C|\geq\exp\bigl(-O(\log^{O(1)}(2K))\bigr)|A|.
$$
Moreover $AC\subseteq A^{O(1)}$, whose size is at most $K^{O(1)}|A|$ by the <higher product bound for an approximate group>. Thus
$$
|AC|\leq\exp\bigl(O(\log^{O(1)}(2K))\bigr)|C|.
$$
The <Ruzsa covering lemma> supplies $X\subseteq A$ of size at most $\exp(O(\log^{O(1)}(2K)))$ such that
$$
A\subseteq XCC^{-1}
=XA_0A_1\cdots A_rA_r\cdots A_1A_0.
$$
Taking the $B_i$ to be the factors in this last product gives
$$
\boxed{A\subseteq XB_1\cdots B_k,}
$$
where $k\leq O(\log^{O(1)}(2K))$ and every $B_i$ is a $K^{O(1)}$-approximate group in $A^{O(1)}$ generating a subgroup of step less than $s$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 2 / b / i / Solution

If $A$ is a finite $K$-<approximate group>, then
$$
A^m\subseteq X^{m-1}A,\qquad |A^m|\leq K^{m-1}|A|
$$
for every <positive integer> $m$, where $A^2\subseteq XA$ and $|X|\leq K$. This follows immediately by <mathematical induction>.

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