I'm reading a book on graph theory, and the book presents the following graph:
They justify that the graph is not Hamiltonian as follows:
Let’s suppose that $G$ is Hamiltonian. Then $G$ contains a Hamiltonian cycle $C$. Since $C$ contains the vertex $t$, which has degree $2$, both $tu$ and $tz$ lie on $C$. By the same reasoning, $xy$ and $xz$ lie on $C$, as do $vw$ and $vz$. However, this says that $z$ is incident with three edges on $C$, which is impossible. Therefore, as we claimed, the graph $G$ is not Hamiltonian.
I think the reason why $tu, tz, xy, xz, vw, vz$ must lie on $C$ is because everytime we go into a vertex, we need to come out? I'm not completely sure about this. Now if these edges must be in the cycle, then I see why $z$ is incident with at least three vertices but not why it's incident with exactly $3$ (why can't $yz$ or $wz$ be on $C$)? Finally, let's suppose $z$ was incident with three edges on $C$. I don't see why this makes it impossible for a Hamiltonian cycle to exist.
Can someone please clarify?