Axiom of Choice:
Given a collection $A$ of nonempty sets, there exists a function $$c: A \to \bigcup_{A_{i} \in A}A_{i}$$ such that $c(A_{i})\in A_{i}$ for all $A_{i} \in A$.
Axiom of Dependent Choice:
Given a nonempty set $A$ and a binary relation $\mathcal{R}$ on $A$ such that for all $a\in A$, there exists $b\in A$ such that $a\mathcal{R}b$. There exists a sequence $$(a_{n})_{n\in \mathbb{N}}$$ such that $a_{n}\mathcal{R}a_{n+1}$ for all $n \in \mathbb{N}$.
Here is my incomplete proof that Axiom of Choice implies Axiom of Dependent Choice:
For $a\in A$, let $R(a)=\{b\in A\mid a\mathcal{R}b\}\implies R(a)\neq\varnothing$ for all $a\in A$.
Using Axiom of Choice for the indexed family of sets $(R(a))_{a\in A}$, there exists a mapping
$$f:A\to A$$ such that $$\forall a\in A:f(a)\in R(a)$$$\text{That is }\forall a\in A:a\mathcal{R}f(a)$. Let $B=\{(a,f(a))\mid a\in A\}$
I don't know how to proceed to prove the existence of the required sequence $(a_{n})_{n\in \mathbb{N}}$ from set $B$.
Please help me complete my proof! Many thanks for you!