For any nonempty set $X$ and any entire binary relation $R$ on $X$, there is a sequence $(x_n)$ in $X$ such that $x_nRx_{n+1}$ for each $n \in \mathbb{N}$. (Here an entire binary relation on $X$ is one such that for each $a$ in $X$ there is a $b$ in $X$ such that $aRb$.)
I cant understand why we cant prove DC bu usual induction: let $x_1=x$ for some $x \in X$ and given $x_n$ there exist $y_n$ such that $x_nRy_n$. We set $x_{n+1}=y_n$. It seems that we dont need AC in proving this theorem.
The reason you can't just prove $\sf DC$ is that you are making infinitely many choices at once. There is, generally, no well-defined mechanism for choosing the specific $x_{n+1}$.
In some cases you can in fact prove that such sequence exist, but those are the exception, not the general rule.
As Andres notes in the comments, $\sf DC$ is vastly weaker than $\sf AC$ itself, although it is quite sufficient for many classical results which need the axiom of choice. Perhaps I should add that your confusion on the topic is exactly why this is such a natural choice principle, and that many people who argued against the axiom of choice were actually using it without knowing.