I am looking at the proof provided by another post:
Prove that if $G$ is a $k-connected$ graph, then $G \bigvee K_1$ is $k+1$ connected graph
I am having trouble understanding the first case how $S-v$ turns out to be a vertex cut for the graph $G.$ I do not see how this could be the case, although I do understand that it would contradict the fact that $G$ is $k$-connected.
Any suggestions?
In the first case for the answer there, $S$ is defined (for the purposes of contradiction) as a vertex cut set of $H:=G+K_1$ of size $k$ that contains $v\in K_1$. Removing $v$ leaves the full graph $G$ and a vertex cut $S-v$ of size $k{-}1$, which is a contradiction, since we are given that $G$ is $k$-connected.
Possibly it's easier to understand as a positive statement: since $v$ from $K_1$ connects to every vertex in $G$, a reduced graph can never be disconnected unless $v$ is removed. Thus any vertex cut set of $H$ must contain the vertex $v$ from $K_1$, and the rest of the cut set clearly requires at least $k$ more vertices due to the $k$-connected nature of $G$.