Prove that if $G$ is a $k$-connected graph, then $G \vee K_1$ is $(k+1)$-connected graph
Here is what I got so far
Since $G$ is a $k-connected$ graph, there exists a vertex cut set $S \subset V(G)$ such that $|S|\geq k$ and $G-S$ is disconnected.
Let $H =G \vee K_1$,for $|S| \leq k$ , $H-S$ is connected since there is still one vertex $v\in K_1$ connected to every vertex $u \in G$ such that $u \not \in S$. I order for $H$ to be disconnected , we need to delete $v$ as well. Since $H- (S \cap \{v\})=G-S$, since $|S| \geq k$, $|S \cap \{v\} \geq k+1$, so $H-(S \cap \{v\})$ is disconnected so $H$ is $k+1$ connected graph.
This i how I understand the problem, but I don't know how to write it formally.
Let $H=G\lor K_1$ and $v$ the vertex of $K_1$ in $H$. We must prove two things:
a) $H$ has at least $k+2$ vertices.
b) $H$ has no vertex cut with a size smaller than $k+1$.
a) is obvious: since $G$ is $k$-connected it has at least $k+1$ vertices, so $H$ has at least $k+2$ vertices.
For b) we argue by contradiction. Suppose $S$ is a vertex cut of size at most $k$.
Case 1: $v\in S$. Then $S-v$ is a vertex cut of $G$ of size at most $k-1$, contradicting the assumption that $G$ is $k$-connected.
Case 2: $v\notin S$. Then every vertex of $H-S$ is either $v$ itself, or connected to $v$, so $H-S$ is connected, contradicting the assumption that $S$ is a vertex cut for H.