Hamiltonian Graph - Prove number of connected components from $G\setminus S$ is $\le |S|$

Question

I failed a graph theory exam last week and I would like to know how to solve some of the problems I got because I don't have any idea. One of the problems is:

Let $G = (V,E)$ be a hamiltonian graph and S is included in $V$. Proof that the number of connected components from $G\setminus S$ is $\le|S|$.

I didn't even had the chance to write something here because I ran out of time. And I don't even know how to start it.. Could you help me with a proof for that, please?

Answer 1

Let $C=v_0,...,v_n,v_0$ be an Hamiltonian cycle in $G$.

Remove the vertices of $S$ from the cycle, you get at most $|S|$ subpath from $C$. Every subpath is included in a connected component of $G\setminus S$, every vertex of $G\setminus S$ is in one of the subpath.

Hence the result.