Let $G=(V,E)$ be a simple graph. Let $r\in V$ be a vertex whose removal (and the removal of its edges) from $G$ will result in graph $G'$ which will have more connected components than $G$. Prove that $G$ does not have Hamiltonian cycle.
Suppose that $G$ is connected and is split into at least two connected components $G_1=(V_1, E_1)$ and $G_2=(V_2, E_2)$. If $|V|=n$ and $|V_1|=p$ then for vertex $a\in V_2$ which was the neighbor of $r$ we have that $\deg(a)\le n-p-1$. For vertex $b\in V_1$ which was the neighbor of $r$ we have that $\deg(b)\le p-1$.
In order for $G$ to have Hamiltonian cycle we must have $\forall x,y \in V:\deg(x)+\deg(y) \ge n$. If we check this for $a$ and $b$: $$ \deg(a)+\deg(b)\le p-1+n-p-1=n-2 $$ Now let's "return" the vertex $r$ to $G'$ such that it becomes $G$ again. Both $a$ and $b$ have one more edge now so: $$ \deg(a)+\deg(b)\le p-1+n-p-1+\mathbf 2=n $$
This still doesn't prove that $G$ doesn't have Hamiltonian cycle but I don't have any other leads.
Proof.
If removing $v\in V(G)$ splits $G$ into several parts, then removing $v$ also split any possibly contained Hamiltonian cycle $C\subseteq G$ ($G$ and $C$ are defined on the same vertex set). But a cycle cannot become disconnected by removing a single vertex, because cycles are $2$-connected.
$\square$
Intuitively, a graph that can be split by a single vertex has a single-vertex "bottleneck" (see the upper graph in the picture below). To have a Hamiltonian cycle however, you need to have bottlenecks of size not smaller than two, because you need to go in both directions through it without using the same vertex twice.