Let $G$ be a graph. If there exists $X\subseteq V(G)$, $X\neq \emptyset$ s.t. $G\setminus X$ has more than $|X|$ components, then $G$ has no Hamiltonian cycle.
Proof. Suppose for a contradiction that $C$ is a Hamiltonian cycle in $G$. Clearly,
$$\text{comp}(C\setminus X)\geq \text{comp}(G\setminus X)>|X|.$$
Let $F$ be the set of edges of $C$ incident to vertices of $X$. Then $|F|\leq 2|X|$. On the other hand, every component of $C\setminus X$ is incident to at least two edges of $F$.
$$|F|\geq 2\text{comp}(C\setminus X),$$
so $\text{comp}(C\setminus X)\leq |X|$, a contradiction.
Why is it that $\text{comp}(C\setminus X)\geq \text{comp}(G\setminus X)>|X|$? The author says "clearly" but this does not seem obvious to me. Could someone help me understand how they saw this and justified it?
$C$ has same vertices as $G$, and edges of $C$ are a subset (possibly equal) of edges of $G$.
So $C\setminus X$ has same vertices as $G\setminus X$, and edges of $C\setminus X$ are a subset (possibly equal) of edges of $G\setminus X$.
If you take a graph and suppress some edges, the number of components can only be greater or equal. So comp$(C\setminus X) \ge$ comp$(G\setminus X)$.
As for comp$(G\setminus X) > |X|$ that's by hypothesis in the demonstration.