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: