Let $A$ be an infinite set and $\mathcal{F}(A)$ be the set of all finite subsets of $A$. Prove that $|A| = |\mathcal{F}(A)|$.
It is trivially true that $|A| \le |\mathcal{F}(A)|$. The idea, I think, is to find an injection from $\mathcal{F}(A)$ to $A$. To this end, let $f: A\to A$ be any function, then $f^{-1}(a) = F$, for $a\in A$, where $F$ is a fiber under $f^{-1}$. Let $(\mathcal{S}, f)$ be the collection of all fibers $F$ under all $f^{-1}$ on $A$ and denote $f(F_i)=a_i$ for all $i\in I$, then $f_i$ is an injection on a subset of $\mathcal{F}(A)$. Also, $\mathcal{F}(A)\subseteq (\mathcal{S},f)$. By the axiom of choice, there exists a choice function $j$ on $(\mathcal{S},f)$ such that $j[(\mathcal{F}(A),f)]\in (\mathcal{F}(A),f)$, for all $F\in \mathcal{F}(A)$. Let $f=\bigcap\limits_{i\in I} f_i$ (defined as the intersection of the domains of all $f_i$), then $f:\mathcal{F}(A)\to A$ is injective. Hence, the conclusion holds.
I'm very new to the axiom of choice and this kind of the set theory, so would appreciate if you could point out any wrongdoings in my proof.