Proving that if $\delta(G)=\min\limits_{v\in V}d(v)$, there exists a path of length $\delta(G)$: confusion about construction of path using neighbors

Question

In a proof of this statement at some point we say that given $v_0v_1\dots v_k$, a path of maximum length, we can suppose that all neighbors of $v_0$ are in the path. I see how we can add one neighbor $u$ to the beginning of the path but if $v$ has more neighbors how does this work?

A similar argument is used in the second part... And I also don't understand how the second part proves the existence of a cycle of length $\delta(G)+1$. To me it only proves that there exists a cycle of length $\ge\delta(G)+1$

This graph has $\delta=3$ but it's not obvious why every neighbor of $4$ or $2$ is in the longest path:

Answer 1

You can use the fact that $v_0, \dots, v_k$ is the path of maximum length to prove the claim about $v_0$'s neighbors. The part where it says "Indeed, assuming that $v_0$ has another neighbor..." is what you're looking for.

As for the second part, the question is probably asking for a cycle of length $\geq$ $\delta(G)+1$. Otherwise, $G=C_n$ would be a counter-example for $ n > 3$.