If a graph is 2-vertex-connected, then it can be produced by $K_3$, using only edge division and addition

Question

I want to prove that if a graph is 2-vertex-connected, then it can be produced by $K_3$ (the simple triangle), using only edge division ("splitting" an edge, $(u,v)$, by creating a new vertex, $x$, deleting the edge $(u,v)$ and adding the edges $(u,x),(x,v)$) and edge addition (just adding any edge).

I want to prove this by induction, if possible. The base case is trivial, as $K_3$ is already 2-vertex-connected.

Let's assume that the above is true for every 2-vertex-connected graph with $|V| < n$. We need to prove that it is true for a graph G with $|V(G)|=n$.

Case 1: $G$ is also 3-vertex-connected

Let $u \in V(G)$. The graph $G \setminus u$ is at least 2-vertex-connected, so by the induction hypothesis it can be produced by $K_3$. The vertex $u$ in $G$ has at least 2 neighbors or else $G$ would not be 2-vertex connected. Let $u_1, u_2 \in G \setminus u$ be 2 of the neighbors of $u$ in G. If the edge $(u_1,u_2)\not\in E(G \setminus u)$, we add this edge and then split it, adding the vertex $u$. If $(u_1,u_2)\in E(G \setminus u)$, we add the edge again. Then, we add all the other edges containing $u$ that previously existed in $G$ and get the original graph $G$.

Case 2: $G$ is not 3-vertex-connected, but there exists a vertex $u$ such that $G \setminus u$ is still 2-vertex-connected

We follow the same procedure as Case 1, this time removing the specific vertex $u$.

Case 3: $G$ is not 3-vertex-connected, $G \setminus v$ is not 2-vertex-connected $\forall v \in V(G)$, but there is at least one vertex $u \in V(G)$ such that $d_G(u)=2$

Let $u_1, u_2 \in N_G(u)$. Since $G$ is 2-vertex-connected, adding the edge $(u_1, u_2)$ does not affect its 2-vertex-connectivity. The graph $G \setminus u$ is obviously still 2-vertex-connected, so by the induction hypothesis it can be produced by $K_3$. Splitting the edge $(u_1, u_2)$ using the vertex $u$ gives us the original $G$.

Case 4: $G$ is not 3-vertex-connected, $\forall v \in V(G), d_G(v) >2$ and $G \setminus v$ is not 2-vertex-connected.

I do not know what to do in that case. I am not sure if such a case can exist, but I cannot prove that it can't. I would appreciate your help.

Answer 1

The following statement is known (here proposition 5 detailed proof):

Let $G$ be a simple $2$-connected graph with no vertex of degree $2$. Then there is a vertex $v$ in $G$ such that $G-v$ is $2$-connected.

It follows that case 4 is impossible.

PS One remark to your text. Apparently a caveat of this kind is required. The graph is not 3-connected, but 2-connected.