= Solution
Assume $A$ is finite and nonempty; the empty case is immediate. Here $2A=A+A$. Choose a nonempty <subset> $X\subseteq A$ minimizing
$$
K=\frac{|X+A|}{|X|}.
$$
Taking $X=A$ as a competitor gives $K\leq C$, and every <subset> $Y\subseteq X$ satisfies $|Y+A|\geq K|Y|$, including $Y=\varnothing$.
We first prove the <Petridis minimal-growth lemma> in the particular form needed: for every <finite set> $S$,
$$
|X+A+S|\leq K|X+S|.
$$
List $S=\{s_1,\ldots,s_t\}$ and let $X_j$ consist of those $x\in X$ for which $x+s_j$ does not belong to an earlier translate $X+s_h$, $h<j$. The <sets> $X_j+s_j$ partition $X+S$, so $|X+S|=\sum_j|X_j|$.
If $x\in X\setminus X_j$, then $x+s_j=x'+s_h$ for some earlier $h$ and $x'\in X$. Hence every point of $x+A+s_j$ already belongs to $X+A+s_h$. Therefore the new points contributed by $(X+A)+s_j$ are contained in
$$
\bigl[(X+A)\setminus((X\setminus X_j)+A)\bigr]+s_j.
$$
Their number is at most
$$
|X+A|-|(X\setminus X_j)+A|\leq K|X|-K|X\setminus X_j|=K|X_j|.
$$
Summing over $j$ proves the lemma. With $S=A$, it gives
$$
|X+2A|\leq K|X+A|=K^2|X|.
$$
We also prove the <Ruzsa triangle inequality> directly. For each $d\in U-V$, choose one representation $d=u_d-v_d$. The map
$$
(d,x)\longmapsto(x+u_d,x+v_d)
$$
from $(U-V)\times X$ into $(X+U)\times(X+V)$ is <injective>: subtracting the output coordinates recovers $d$, after which either coordinate recovers $x$. Thus $|U-V||X|\leq|X+U||X+V|$. Taking $U=V=2A$ now yields
$$
\boxed{|2A-2A|\leq\frac{|X+2A|^2}{|X|}\leq K^4|X|\leq C^4|A|.}
$$
This proves the required growth and <difference set> estimates directly.
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