Please have a look at below proof! Thank you for your help so much
- Axiom of Dependent Choice
Let $T \neq\varnothing$ and $\mathcal{R} \subseteq T^2$ such that $\forall a \in T, \exists b \in T: a\mathcal{R}b$. Then there exists $(x_n \mid n \in \mathbb N)$ such that $x_n \mathcal{R} x_{n+1}$.
- Axiom of Countable Choice
Let $(A_n \mid n \in \mathbb N)$ be a sequence of non-empty sets and $X=\bigcup_{n \in \mathbb N} A_n$. Then there exists a mapping $f: \mathbb N \to X$ such that $f(n) \in A_n$.
My proof that Axiom of Dependent Choice implies Axiom of Countable Choice:
Let $\mathcal{R}=\{(a,b) \in X^2 \mid \exists n \in \mathbb N, a \in A_n \text{ and } b \in A_{n+1}\}$.
$\mathcal{R}$ satisfies the requirement of Axiom of Dependent Choice. Hence there exists a sequence $(x_i \mid i \in \mathbb N)$ such that $x_i \mathcal{R} x_{i+1}$ where for some $n$, $x_i \in A_{i+n}$ for all $i \in \mathbb N$.
Let $f:\mathbb N \to X$ such that $f(i) \in A(i)$ for all $i<n$ and $f(i)=x_{i-n}$ for all $i \geq n$.
Defining $f$ in this way, we get a function as desired.
PS: From @spaceisdarkgreen's answer,I fixed $\mathcal{R}$ as follows
Let $Y=\{(a,i) \mid a \in A_i\}$ and $\mathcal{R}=\{((a,m),(b,n)) \in Y^2 \mid n=m+1\}$.
This is on the right track, but there is an error.
The problem is your relation does not preclude the possibility that, say $x_1\in A_{52}$ and $x_2\in A_{53},$ but also $x_2\in A_{103}$ and $x_3\in A_{104},$ all while $x_3\notin A_{54}.$ So it is not necessarily true that there is an $n$ such that $x_i\in A_{i+n}$ for all $i.$
One trick for fixing it is would be to start by rigging the $A_i$ to be disjoint by replacing the elements $a\in A_i$ with tagged elements $\langle i,a\rangle.$