[proofplan] Choose $X \subset A$ maximal subject to the translates $x+B$, with $x \in X$, being pairwise disjoint. The disjoint union of these translates lies inside $A+B$, so the small-doubling hypothesis bounds $|X|$. Maximality then forces every translate $a+B$ with $a \in A$ to intersect one of the selected translates, and that intersection identity rearranges to $a \in X+B-B$. [/proofplan] [step:Choose a maximal disjoint family of $B$-translates indexed by elements of $A$] Let $\mathcal{F}$ be the finite collection of subsets $Y \subset A$ such that the family of translates \begin{align*} \{y+B : y \in Y\} \end{align*} is pairwise disjoint. Since $\varnothing \in \mathcal{F}$ and $A$ is finite, there exists $X \in \mathcal{F}$ with maximal cardinality. Thus $X \subset A$, the translates $x+B$ for $x \in X$ are pairwise disjoint, and no strictly larger subset of $A$ has this disjointness property. [guided] We need a small set $X$ whose $B-B$-neighbourhood covers $A$. The standard way to build it is to select as many disjoint translates of $B$ as possible, with translating elements taken from $A$. Define $\mathcal{F}$ to be the finite collection of all subsets $Y \subset A$ such that the family \begin{align*} \{y+B : y \in Y\} \end{align*} is pairwise disjoint. This collection is nonempty because $\varnothing \in \mathcal{F}$. It is finite because $A$ is finite, so among all elements of $\mathcal{F}$ there is one with maximal cardinality. Choose such a subset and call it $X$. By construction, $X \subset A$, the translates $x+B$ for $x \in X$ are pairwise disjoint, and maximality means the following precise statement: if $a \in A \setminus X$, then the enlarged family indexed by $X \cup \{a\}$ is not pairwise disjoint. [/guided] [/step] [step:Bound the size of the maximal set using the small sumset hypothesis] For every $x \in X$, since $x \in A$, we have \begin{align*} x+B \subset A+B. \end{align*} The translates $x+B$ are pairwise disjoint and each has cardinality $|B|$, because the map $B \to x+B$ given by $b \mapsto x+b$ is a bijection. Therefore \begin{align*} |X|\,|B| = \left|\bigcup_{x \in X} (x+B)\right| \leq |A+B| \leq K|B|. \end{align*} Since $B$ is nonempty, $|B|>0$, and division by $|B|$ gives $|X| \leq K$. [guided] The purpose of maximal disjointness is that it converts the size of $X$ into a counting estimate. For each $x \in X$, the translate $x+B$ lies in $A+B$ because $x \in A$ and every element of $x+B$ has the form $x+b$ with $b \in B$. Also, translation by $x$ is a bijection from $B$ to $x+B$: \begin{align*} B &\to x+B \\ b &\mapsto x+b. \end{align*} Its inverse is $x+B \to B$, $z \mapsto z-x$. Hence $|x+B|=|B|$ for every $x \in X$. Since the sets $x+B$ are pairwise disjoint, the cardinality of their union is the sum of their cardinalities: \begin{align*} \left|\bigcup_{x \in X} (x+B)\right| = \sum_{x \in X} |x+B| = |X|\,|B|. \end{align*} Because this union is contained in $A+B$, and because the hypothesis gives $|A+B| \leq K|B|$, we obtain \begin{align*} |X|\,|B| \leq |A+B| \leq K|B|. \end{align*} The set $B$ is nonempty, so $|B|>0$. Dividing by $|B|$ yields $|X| \leq K$. [/guided] [/step] [step:Use maximality to force every translate $a+B$ to meet a selected translate] Let $a \in A$ be arbitrary. If $a \in X$, then $a \in X+B-B$ because $B$ is nonempty: choosing any $b_0 \in B$ gives \begin{align*} a = a+b_0-b_0 \in X+B-B. \end{align*} Now suppose $a \in A \setminus X$. By maximality of $X$, the family of translates indexed by $X \cup \{a\}$ is not pairwise disjoint. Since the translates indexed by $X$ are pairwise disjoint, there exists $x \in X$ such that \begin{align*} (a+B) \cap (x+B) \neq \varnothing. \end{align*} Choose $b_1,b_2 \in B$ with \begin{align*} a+b_1 = x+b_2. \end{align*} Rearranging in the abelian group $G$ gives \begin{align*} a = x+b_2-b_1 \in X+B-B. \end{align*} Thus every $a \in A$ belongs to $X+B-B$, and hence \begin{align*} A \subset X+B-B. \end{align*} [guided] We now turn maximality into covering. Fix an arbitrary element $a \in A$. We must prove $a \in X+B-B$. First consider the case $a \in X$. Since $B$ is nonempty, choose $b_0 \in B$. Then \begin{align*} a = a+b_0-b_0, \end{align*} with $a \in X$ and $b_0,b_0 \in B$, so $a \in X+B-B$. Now consider the case $a \in A \setminus X$. Because $X$ was chosen maximal among subsets of $A$ whose $B$-translates are pairwise disjoint, the enlarged set $X \cup \{a\}$ cannot still have pairwise disjoint translates. The old translates $x+B$ with $x \in X$ are already pairwise disjoint, so the only possible failure is that the new translate $a+B$ intersects one of them. Therefore there exists $x \in X$ such that \begin{align*} (a+B) \cap (x+B) \neq \varnothing. \end{align*} Choose an element in this intersection. By the definition of the two translates, there exist $b_1,b_2 \in B$ such that \begin{align*} a+b_1 = x+b_2. \end{align*} Subtracting $b_1$ from both sides in the abelian group gives \begin{align*} a = x+b_2-b_1. \end{align*} Here $x \in X$ and $b_2,b_1 \in B$, so $a \in X+B-B$. Since the element $a \in A$ was arbitrary, this proves \begin{align*} A \subset X+B-B. \end{align*} [/guided] [/step] [step:Combine the cardinality estimate and the covering inclusion] The set $X \subset A$ constructed above satisfies both \begin{align*} |X| \leq K \end{align*} and \begin{align*} A \subset X+B-B. \end{align*} This is exactly the asserted conclusion. [/step]