Under the ultrafilter lemma, abelian groups are amenable

Suppose first that $A$ is finitely generated. By [L1], write $A\cong {\mathbb{Z}}^r\oplus T$ with $T$ finite. Let $S\subseteq A$ be finite and let $\epsilon >0$. If $r=0$, then $A=T$ is finite and $F=A$ satisfies $sF=F$ for every $s\in S$, so the Folner condition is immediate. Assume now that $r\ge 1$. Transport $S$ across the isomorphism, and let $M$ be the maximum of the ${\mathrm{\ell }}^{\mathrm{\infty }}$-norms of the ${\mathbb{Z}}^r$-components of the transported elements. For $n\ge 1$, put $B_n=[-n,n{]}^r\times T$. Then every translate by an element of $S$ changes only the $M$-thick boundary layers of the box, so $\mathrm{\mid }(s+B_n)\mathrm{△}{B}_n\mathrm{\mid }=O(n^{r-1})$ uniformly in $s\in S$, while $\mathrm{\mid }{B}_n\mathrm{\mid }=(2n+1{)}^r\mathrm{\mid }T\mathrm{\mid }$. For large $n$ this gives $\mathrm{\mid }(s+B_n)\mathrm{△}{B}_n\mathrm{\mid }<\epsilon \mathrm{\mid }{B}_n\mathrm{\mid }$ for every $s\in S$. Thus finitely generated abelian groups satisfy the Folner condition.

By [L2], every finitely generated abelian group is therefore amenable.

Now let $A$ be arbitrary. Given a finite subset $S\subseteq A$ and $\epsilon >0$, the subgroup $⟨S⟩$ is finitely generated and abelian, so step 2.1 makes it amenable. Applying [L2] inside $⟨S⟩$ yields a finite nonempty set $F\subseteq ⟨S⟩$ with $\mathrm{\mid }sF\mathrm{△}F\mathrm{\mid }<\epsilon \mathrm{\mid }F\mathrm{\mid }$ for every $s\in S$. The same set $F$ witnesses the Folner condition in $A$. Since $S$ and $\epsilon$ were arbitrary, [L2] shows that every abelian group is amenable.