Graph theory proof - if $G$ is $2$-connected then for each $u$ there exists $v$ and a cycle $C$ that includes $u, v$ and all of $v$'s neighbors

Question

Need to prove that if $G$ is $2$-connected then then for each $u$ there exists a $v$ and a cycle $C$ that includes $u$, $v$ and all of $v$'s neighbors. I have reached an insight that all vertices must be of degree 3 or more besides one (otherwise, the proof is trivial). In addition to that, I have tried taking the cycle with the most chords that contains $u$, but seemed to have no real progress with it.

Answer 1

Theorem: In a 2-connected graph $G$, for every vertex $u$, there exists a different vertex $v$ and a cycle $C$ that contains $u$, $v$, and all the neighbors of $v$.

Lemma: In a connected graph where each vertex has a degree of at least 2, there exists a vertex $v$ and a cycle $C$ that contains $v$ and all of $v$'s neighbors.

Proof of Lemma: Consider a connected graph $G$ where each vertex has a degree of at least 2. Let $P$ be the longest path in $G$, denoted as $P = v_0 - v_1 - \ldots - v_k$. Since every vertex in $G$ has a degree of at least 2, the endpoint $v_k$ must have at least two neighbors. These neighbors must be on the path $P$; otherwise, connecting $v_k$ to a neighbor that is not already on the path would contradict the assumption that $P$ is the longest path. Let $v_i$ be the neighbor of $v_k$ on $P$ with the smallest index (so there is no neighbor $v_j$ such that $j<i$). Construct a cycle by starting from $v_i$, following $P$ to $v_k$, and then returning to $v_i$. This cycle includes $v_k$ and all its neighbors on $P$, as $v_i$ has the smallest index among these neighbors. $\blacksquare$

Proof of Theorem: Let G be a 2-connected graph and u be a vertex in it. Assume all other vertices have a degree of at least 3. If any vertex $v$ different from $u$ has a degree of 2, then by Menger's theorem, there is a cycle containing both $u$ and $v$, and since $v$ has only two neighbors, this cycle must pass through both neighbors. Now, remove $u$ from $G$, creating a subgraph $G'$. In $G'$, each vertex still has a degree of at least 2, so by the lemma, there is a vertex $v$ and a cycle $C'$ containing $v$ and all of $v$'s neighbors. Reintroduce $u$ to $G$. There must be a path $P_1$ from $u$ to a vertex on $C'$ such that $P_1$ only intersects $C'$ at the endpoint (the proof is trivial. Let $x$ be a vertex in $G$. Since $G$ is 2-connected, there must be a path $P$ from $u$ to $x$. If $P \cap C' = \{ x \}$ then we are done and $P_1 = P$. If there are more intersection points brtween $P$ and $C'$, then we choose the endpoint of $P_1$ to be the first intersection point). Let $C' = v_0 - v_1 - \ldots - v_k - v_0$ where $v_0$ is the endpoint of $P_1$. Consider the path $P' = u - P_1 - v_0 - v_1 - \ldots - v_k$. Since $G$ is 2-connected, by Menger's theorem, there is another path $P_2$ from $u$ to $v_k$ that is disjoint from $P'$. Construct the cycle $C = u - P1 - v_0 - v_1 - \ldots - v_k - P_2 - u$. This cycle includes $u$, travels to the cycle $C'$ using $P_1$, traverses through all the vertices in $C'$ ($v_0-v_1-\ldots-v_k$) including $v$ and all its neighbors, and returns to $u$ via $P_2$ closing the cycle, thus proving the theorem. $\blacksquare$