I would like to discuss the Ruzsa's covering lemma from Tao-Vu book. The proof given in the book skips some details and I've decided to write all the details just to be sure that I understand everyhting in a correct way. Please read my proof and let me know is everything correct there.
Lemma 2.14 (Ruzsa's covering lemma) For any additive sets $A, B$ with common ambient group $Z$, there exists an additive set $X_+\subseteq B$ with $$B\subseteq A-A+X_+; \quad |X_+|\leq \dfrac{|A+B|}{|A|}; \quad |A+X_+|=|A||X_+|$$ and similarly there exists an additive set $X_-\subseteq B$ with $$B\subseteq A-A+X_-; \quad |X_-|\leq \dfrac{|A-B|}{|A|}; \quad |A-X_-|=|A||X_-|.$$ In particular, $B$ can be covered by $\min (\frac{|A+B|}{|A|}, \frac{|A-B|}{|A|})$ translates of $A-A$.
Proof:
I) We'll prove an existence of an additive set $X_+\subseteq B$ with desired properties. Consider the following set: $$\mathcal{C}:=\{X\subseteq B: X\neq \varnothing \ \text{and} \ (A+x)\cap (A+x')=\varnothing \ \forall x,x'\in X \ \text{such that} \ x\neq x'\}.$$ Since $B$ is an additive set, then $\mathcal{C}\neq \varnothing$ because $\{b\}\in \mathcal{C}$ for each $b\in B$.
We note that $\mathcal{C}$ is partially ordered set (by inclusion). Suppose that $T$ is a totally ordered subset of $\mathcal{C}$, then $T$ has an upper bound in $\mathcal{C}$.
Indeed, if $T=\{X_i\}_{i=1}^N,$ then consider an element $L:=\cup_{i=1}^N X_i$. It is obvious that $X_i\subseteq L$ for any $1\leq i\leq N$ (that means that $L$ is an upper bound for $T$) and also we'll prove $L\in \mathcal{C}$.
It is easy to see that $L\subseteq B$. Suppose that $t_1,t_2\in L$ such that $t_1\neq t_2$, then $t_1\in X_{i_1}$ and $t_2\in X_{i_2}$. Since $T$ is totally ordered, then WLOG assume that $X_{i_1}\subseteq X_{i_2}$. Hence $t_1,t_2\in X_{i_2}$ and since $X_{i_2}\in \mathcal{C}$, then $(A+t_1)\cap (A+t_2)=\varnothing$. It proves that $L\in \mathcal{C}$.
We were able to show that $\mathcal{C}$ is a partially ordered set and any chain in $\mathcal{C}$ has an upper bound in $\mathcal{C}$, then by Zorn's lemma $\mathcal{C}$ has a maximal element $X_+$. Since $X_+\neq \varnothing$ (because $X_+\in \mathcal{C}$ and elements of $\mathcal{C}$ are nonempty by construction), then $X+\subseteq B$ is an additive set.
i) we see that $A+X_+\subseteq A+B$, then $|A+X_+|\leq |A+B|.$ However, $A+X_+=\sqcup_{t\in X_+}(A+t)$ and the last union should be disjoint because $X_+\in \mathcal{C}$. Hence $$|A+X_+|=|\sqcup_{t\in X_+}(A+t)|=\sum \limits_{t\in X_+}|A+t|=\sum \limits_{t\in X_+}|A|=|A||X_+|\leq |A+B| \Rightarrow$$ $$|X_+|\leq \frac{|A+B|}{|A|} \ \text{and} \ |A+X_+|=|A||X_+|.$$
ii) If $X_+=B$, then obviously $B\subseteq A-A+X_+$.
Suppose that $X_+\subsetneq B.$ If $b\in X_+$, then obviously $b\in A-A+X_+.$ If $b\in B\setminus X_+,$ then $(A+b)\cap (A+x)\neq \varnothing$ for some $x\in X_+$. Hence $\exists \alpha\in A+b$ and $\alpha \in A+x$, i.e. $\alpha=a_1+b=a_2+x$ for some $a_1,a_2\in A$, then $b=a_2-a_1+x\in A-A+X_+$. Therefore, $$B\subseteq A-A+X_+.$$
Remark 1: The proof of an existence of $X_-$ follows if we apply part I) to the pair of additive sets $A$ and $-B$.
Remark 2: I am more than sure that there is a typo in the last line of statement of the lemma. It should be $B$ can be covered by $\leq\min (\frac{|A+B|}{|A|}, \frac{|A-B|}{|A|})$ translates of $A-A$.
Please let me know is my reasoning correct and what do you think about Remark 2?