Why does each subfield have to contain multiples of 1? I have seen a Theorem:

every field contains exactly one prime subfield $K_0$.

and next:

$K_0$ has to contain 1 and all its multiples: $n \cdot 1 = 1 + \ldots + 1$

Is this because it has to contain 0 and 1 by definition, and then it must also holds that $1 + 1$ belongs to K, and $(1 + 1) + 1$ too because of two-arguments $+$ operation of its abelian group?
so we can prove it by induction?

Yes, that is correct.
The base case ($n=1$) is trivial.
For the inductive step let $n\in \Bbb N$ such that $n\cdot 1_K \in K$.
By definition we have $(n+1)\cdot 1_K=n\cdot 1_K + 1_K$. Since $n\cdot 1_K\in K$ (by our hypothesis) and since $1_K\in K$, from the fact that $K$ is a field it follows that $n\cdot 1_K + 1_K\in K$, that is  $(n+1)\cdot 1_K\in K$, which completes the induction.
Edit: The fact that $K$ with $+$ is abelian is irrelevant. You just use the fact that it is a group.
