Combinatorics Lemma for Brooks' Theorem I have the following lemma for Brooks' Theorem I a trying to understand:
Lemma: Let $G$ be a $2$-connected graph with $\delta(G) \geq 3$. If $G$ is not complete, then $G$ contains an induced path on 3 vertices, say $uvw$, such that $G\setminus \{u,w\}$ is connected.
This is supposed to help in proving Brooks' Theorem in the case where $\Delta(G) \geq 3$, so the appearance of $\delta(G) \geq 3$ in the lemma seems very odd to me. In addition, the proof given doesn't make sense to me yet.
Proof: Since $G$ is connected and not complete, it contains an induced path on $3$ vertices. If $G$ is $3$-connected, any such path will do. Otherwise, let $\{v,x\} \subset V(G)$ be a cutset. Since $G−v$ is not $2$-connected, it has at least two endblocks $B_1, B_2$. Since $G$ is $2$-connected, each endblock of $G−v$ has a noncutvertex adjacent to $v$. Let $u\in B_1$ and $w\in B_2$ be such vertices.  Now $G\setminus \{u,w\}$ is connected since $d(v)\geq3$. So $uvw$ is our desired induced path.
What I think this is saying is that we can first find three vertices $u,v,w$ so that $uv$ and $vw$ are edges while $uw$ is not an edge. Now if $G$ is $3$-connected then we can always remove two vertices and stay connected, so we take $G$ to be exactly $2$-connected. We can take $\{v,x\}$ to be a cutset, so removing just one from $G$ yields the graph $G-v$ where $x$ is a cut vertex. Since $G$ has a cutvertex it has (at least) two connected components $B_1, B_2$. I don't see why each $B_i$ has a noncutvertex adjacent to $v$ however. I get the rest of the proof, but it seems odd to impose $\delta(G) \geq 3$ since that is not a condition in Brooks' Theorem.
The statement and its proof were taken from here
 A: Here is a proof of the lemma. It's not pretty; I'm sure a real graph theorist could do it much better.
Suppose $G-x-v$ is disconnected, with components $A_1,A_2,\dots,A_k$, where $k\ge2$. Since $G-v$ and $G-x$ are connected graphs, both $x$ and $v$ have at least one neighbor in each component $A_i$.
Claim. We can choose vertices $u\in N(v)\cap A_1$ and $w\in N(v)\cap A_2$ so that $x$ is connected to $v$ in the graph $H=G-u-w$. (That means connected by a path, not necessarily an edge; in other words, $x$ and $v$ are in the same component of $H$.)
Case 1. $k\ge3$.
Now each of the vertices $x$ and $v$ has a neighbor in $A_3$, which is connected; so we can choose any vertices $u\in N(v)\cap A_1$ and $w\in N(v)\cap A_2$.
Case 2. $k=2$.
If $xv\in E(G)$ there is nothing to prove, so we assume $xv\notin E(G)$. Since $\delta(G)\ge3$, $v$ has at least two neighbors in some $A_i$; we may assume that $v$ has two neighbors $u',u''$ in $A_1$. Choose $y\in N(x)\cap A_1$. We may assume that $d(y,u')\le d(y,u'')$; thus there is a path from $y$ to $u'$ in $A_1$ which does not go through $u''$. Let $u=u''$ and choose a vertex $w\in N(v)\cap A_2$. Then $x$ is connected to $v$ in $H=G-u-w$, via $y$ and $u'$.
Now suppose $u$ and $w$ have been chosen according to the Claim, and let $S$ be the component of $H=G-u-w$ containing $x$ and $v$; we have to show that $S=V(H)$. Consider any vertex $z\in V(H)\setminus\{x,v\}$; we may assume that $z\notin A_2$. Since $G-u$ is connected, there is a path from $\{x,v\}$ to $z$ in $G-u$. If $P$ is the shortest such path then $P$ contains no vertex of $A_2$ and in particular does not go through $w$, so $P$ is a path from $\{x,v\}$ to $z$ in $H$, showing that $z\in S$.
