Simple $2$-connected Graph with $\chi(G)=3$ I need to prove that for $G$, a simple $2$-connected graph with chromatic number $\chi(G)=3$,
that every $v \in V(G)$ is contained in an odd cycle.
Something tells me I need to somehow show that every $v$ is contained in a subgraph with a minimal degree $2$ and then apply Brook's theorem, but I have not been able to show it in this case.
Any guidance would be appreciated!
 A: The problem is a result of the following proposition, since a $3$-chromatic graph must contain an odd cycle
Proposition: Suppose a $2$-connected graph has an odd cycle. Then there is an odd cycle through each vertex.
Proof: Let $C$ be an odd cycle in a $2$-connected graph $G$. For any vertex $u\in G\backslash C$, there exists vertex-disjoint paths $P_1$ and $P_2$ from $u$ to two vertices of $C$, say $v_1$ and $v_2$. This is a consequence$^1$ of Menger's theorem and the $2$-connectedness of the graph. Let $C_1$ and $C_2$ be the two halves of the cycle $C$ separated by $v_1$ and $v_2$. The lengths of $C_1$ and $C_2$ have different parities since $C$ is odd. Therefore one of 
$$P_1 + C_1 + P_2$$
or
$$P_1 + C_2 + P_2$$
is a cycle of odd length containing $u$. $\square$
$^1$ There is a very useful consequence of Menger's theorem, which is the concept of a $k$-fan.
Definition: Let $G$ be a $k$-connected graph. Let $v$ be a vertex in $G$ and $S$ a set of $k$ or more vertices in $G\backslash\{v\}$. Then a set of $k$ paths from $v$ to $S$ is called a $k$-fan if the paths intersect pairwise only at $v$. We also call such a fan a $(v,S)$-fan.
Theorem: Let $G$ be $k$-connected. For any $v\in G$, and any set $S$ of $k$ or more vertices in $G\backslash \{v\}$, there exists a $(v,S)$-fan.
Proof: Adjoin a new vertex $u$ to $G$ by connecting $u$ to each element of $S$. Since $|S|\ge k$, it follows that the new graph $G\cup \{u\}$ is also $k$-connected. Menger's theorem then asserts that there exists $k$ vertex-independent paths from $v$ to $u$. Removing $u$ leaves the desired $(v,S)$-fan. $\square$
What we've used in our proposition is the existence of a $2$-fan from $u$ to $C$.
A: Let $B$ represent a subgraph of $G$ such that $\forall v' \in B$, $v'$ is in an even cycle. If $\exists B = G\implies \chi(B) = \chi(G) = 2 \implies B \neq G$
$|V(G) \cap V(B)| = 1 \implies \exists v\in V(G) $such that $v$ is a cut vertex $\implies |V(G)\cap V(B)|\geq 2 $
The proof thus follows:
The remaining vertices $R = V(G)-V(B)$ must connect to opposite sides of the bipartite graph with an odd number of vertices or to the same side of the  bipartite graph with an even number of vertices. Otherwise we have a graph $B'$ which is $B$ + vertices which also satisfy the properties of a bipartite graph. We know that there must be at least one such vertex $v \notin B'$, as $\chi (G) = 3$.
Adding any vertex in $R$ must create an odd cycle: There are $|V(B)|$ connected elements, and as $B$ is an even cycle, $|V(B)| + V(R'_{odd})$ connected elements if you add in vertices which link different sides of the graph $\implies$ all vertices are in an odd cycle. In the case in which you add elements which connect the same side of the graph, your original path contains an odd number of vertices to which you add an even number of vertices which clearly creates an odd cycle. As the graph is 2-connected, we know that splitting the graph in this manner creates another cycle of size $|B| + |R'| - |C_1|$  which is an even number minus an odd number $\implies C_1, C_2$ are both odd. As all remaining vertices in $R$ form an odd cycle in a proper construction of $B$, we have shown that all vertices must be contained in an odd cycle.
