Disprove for every $k$ vertices of $G$ lies on a common cycle of $G$ for $k \geq 2$, $G$ is $k$ connected graph
I want to disporve this so I'm trying to show that $G$ is not a$$ connected graph.
I'm looking at the cycle. Let $G = C_n$ for $n \geq k \geq 3$ because the smallest cycle has 3 vertices. In $C_n$ every vertex lies on a big common cycle. Now I need to consider 2 cases
Case 1: $n =k$, then if I delete $k$ vertices in $C_n$ , I have a null graph, which have nothing to prove.
Case 2: $n>k$ then $n \geq k+1>3$. Let $S=\{v_1, v_2,v_3, \dots , v_k\}$ such that $v_i$ adjacent to $v_{i+1}$, then $C_n -S=P_{n-k}$ which still connected. So this doesn't work for every $k$ vertices in $C_n$, thus the statement is also false in this case.
I'm not sure if my reasoning is acceptable in case 2? I feel like I've missed something
In order to disprove this general statement, you just need one completely specific counterexample.
Consider $C_5$. This satisfies the hypothesis for $k=3$ but is not $3$-connected. This disproves the claim.