Prove that if a graph $G$ has a Hamilton path, then for every $S\subseteq V(G)$, the number of components of $G-S$ is at most $|S|+1$.
My solution (rough and incorrect): Consider a Hamilton path $P$ in $G$. $P$ has to visit each of the components of $G-S$. There is no direct path between two components so the path $P$ has to go to $S$ when it leaves a component. So the number of components in $G-S$ is at most $|S|$.
This is obviously incorrect. What am I missing here? How would it be $|S|+1$ and not $|S|$. Also how can I show that the arrival vertices are distinct? Surely if you prove it this way, you could have just, for example, one vertex in $S$ which has an edge connecting to all of the components.