Prove that in a $2-$connected graph, every two vertices belong to a common cycle. You answered this:The steps followed in the proof are as under:$-$

Base case:$-$ For a cycle graph $,$ any two vertices are connected by the cycle itself, which is a common cycle.

Inductive step:$-$ Lets assume that the claim holds for all $-$connected graphs with fewer than $-$vertices.

Consider a $-$connected graph with $-$vertices. Let and be two vertices in $.$ Since is $-$connected. By property of $-$connected graphs, there exists a cycle containing both and $.$

Now by shortest path analysis considering the following two cases:$-$

if the shortest path between and is a single edge, then and are adjacent, and the edge itself forms a cycle, satisfying the claim.

Multi$-$edge path:$-$ Here is the crucial part:$-$

If contains multiple edges, consider the graph obtained from by deleting all the edges of $.$ remains $-$connected because removing a path does not disconnect a $-$connected graph.

By the induction hypothesis, any two vertices in belong to a common cycle. Since is a shortest path between and $,$ adding it back forms a cycle containing both and $.$

Thus, any two vertices in belong to a common cycle.

However,you should build induction on e$.$g$.$ number of e dges, vertex degree etc, not a structure of gra ph G$.$ That is inconvenient approach. Starting with a cycle $-$ you assume too much about suc h a graph. There are $2-$connected graphs they do $*$not$*$ contain C$\_$n$.$

Regarding inductive step: case a$)$ is wrong. Sin gle edge doesn't form a cycle!

b$)$ deleting an $*$edge$*$ in vertex $2-$connected gr aph proves nothing. You should delete a $*$verte x$*.$

I$\text{'}$m looking forward