For all graphs $G$, if $G$ is semi-Hamiltonian (contains a Hamilton path, i.e., a path that visits every vertex in $G$), then removing any $k$ vertices results in a graph with at most $k+1$ connected components.
Here is my solution.
$\textit{Proof.}$ Assume $G$ is semi-Hamiltonian. Then $G$ contains a path that visits every vertex in $G$. Thus, $G$ is connected. Pick $k$ vertices in $G$. If none of the $k$ vertices are bridges (That is, vertices whose removal results in a disconnected graph), then $G$ contains only 1 connected component, itself.
If all of the $k$ vertices are bridges, then if we remove the first we get $2$ connected components, then we remove the next vertex to get $3$ connected components and so on until we have removed all $k$ vertices and then have $k+1$ connected components.
If $m < k$ of the vertices were bridges then we will have $m+1 < k+1$ connected components.
To me this sounds fine but could anyone say if this sounds good or not to see if I make any leaps that were not able to be made? Thanks very much I appreciate it!