Proof that if a biconnected component has AT LEAST ONE odd-length simple cycle, then ALL edges in such component are in some odd-length simple cycle? Is it true that if a biconnected component has at least one odd-length simple cycle, then ALL edges in such component belong to some (not necessarily the same) odd-length simple cycle? If that's the case, what would be a formal proof for that?
 A: Yes, that statement is true. 
The main idea is that
$$G\text{ has no odd-length simple cycle}
\quad\Leftrightarrow\quad G\text{ is bipartite},$$
which comes up fairly often.
Recall that $G$ is bipartite if
there is a ''coloring'' of vertices $\chi : V(G) \to \{0,1\}$
such that every edge in $G$ joins vertices of opposite color.
The implication $(\Rightarrow)$ holds because, assuming no odd cycles, 
we get a bipartite coloring using the naive/greedy algorithm.
Here is an argument:
Let $e $ be an edge with endpoints $x,y$ in a biconnected graph $G$. 
Suppose that $e$ does not belong in any odd-length simple cycle.
Note that $C$ is a simple cycle containing $e$ if and only if
$C \backslash e$ is a simple path connecting $x$ to $y$
in the deleted graph $G \backslash e$.
Thus our assumption means that in $G\backslash e$,
every simple path from $x$ to $y$ has an odd number of edges.
Additionally, the assumption that $G$ is biconnected implies that 
$$ \text{every edge of $G\backslash e$ lies on a simple path  in $G\backslash e$ from $x$ to $y$}. $$
(This is a common result in graph theory / matroid theory, but I don't remember a reference off hand.)
This means we can assign a $2$-coloring of the vertices in $G\backslash e$,
by alternating $0$'s and $1$'s along paths from $x$ to $y$.
This coloring $\chi : V(G\backslash e) \to \{0,1\}$ satifies
$$ \chi(v) = \begin{cases}
0 & \text{if $v$ is even distance from } x \quad(\text{odd dist. from } y)\\
1 & \text{if $v$ is odd distance from } x \quad(\text{even dist. from } y).
\end{cases}$$
In particular, $x$ and $y$ have opposite colors so this gives a bipartite coloring of $G$.
This shows  that for biconnected $G$,
$$ e\in E(G) \text{ belongs in no odd-length simple cycle} 
\quad\Rightarrow\quad G \text{ is bipartite}. $$
In particular, $G$ cannot have an odd-length cycle.
