I believe the question is resolved to the satisfaction of the OP. See the comments and the revisions to the question for the relevant discussions.$\newcommand{\len}{\operatorname{len}}$
Here I present a different, and--in my mind--conceptually cleaner proof of the same fact.
Assume $G$ is a connected graph such that all of whose cycles are of even length. We generalize this slightly to the following
Proposition. Any closed walk in $G$ has even length.
Proof. Towards a contradiction, suppose not. Let $W$ be a closed walk of odd length such that the length of $W$ is as small as possible. By hypothesis, $W$ cannot be a cycle; i.e., $W$ visits some intermediate vertex at least twice. Hence we can write $W$ as the "concatenation" of two non-trivial closed walks $W_1$ and $W_2$, each of which is shorter than $W$. Further, $\len W_1 + \len W_2 = \len W$, which is odd. Thus at least one of $W_1$ and $W_2$ is of odd length, contradicting the minimality of $W$. Thus there cannot be any closed walk in $G$ of odd length. $\quad\quad \Box$
Partitioning the graph into even and odd vertices. Now, fix a vertex $v$, and define $E$ (resp. $O$) be the set of vertices $x$ in $G$ such that there is an even-length (resp. odd-length) walk from $v$ to $x$. The sets $E$ and $O$ partition $V$:
Assuming $G$ is connected, then clearly $E \cup O = V$.
We now show that $E \cap O = \emptyset$. To the contrary, suppose $x$ is in both $E$ and $O$. Then there is a $v$-$x$ walk $W_1$ of even length and another one $W_2$ of odd length. Then the walk $W_1 \circ \operatorname{reverse} (W_2)$ is a closed walk in $G$ of odd length, a contradiction.
Finally, we show that every edge crosses the cut $(E, O)$:
Assume $x \in E$ and $xy$ is an edge. Then there exists a $v$-$x$ walk $W$ of even length. Therefore, $W \circ xy$ is a $v$-$y$ walk and it has odd length. Therefore, $y \in O$.
Similarly, if $x \in O$ and $xy$ is an edge, we can show that $y$ is in $E$. This proof is similar to the above case.
This establishes that $G$ is bipartite, as desired.
Since you've drawn your example in the plane, cycles contain faces, so we can readily check it's correct: it has exactly two $3$-cycles and exactly one $4$-cycle. Any other cycle contains some faces, and must have length greater than $4$.
I ran a Nauty script to generate the 6-vertex 8-edge graphs, and manually checked which have exactly two 3-cycles and exactly one 4-cycle. These are the three possibilities up to isomorphism:
Yours is the example in the middle.
In the first two examples, we have faces surrounded by 3, 3, 4, and 5 edges. Combining any two faces gives a cycle of length 5 or more.
In the third example, we have faces surrounded by 3, 3, 5, and 5 edges. There's a 4-cycle around the combined 3-edge faces. Combining faces in any other way, gives a cycle of length 5 or more.
Best Answer
Hint: Draw a cycle of length 9. Can you see how to build a 6-cycle and a 7-cycle with the extra vertex? Now remains to build the 5-cycle and the 8-cycle. You can build the 5-cycle by adding only 1 edge. I hope this can help you without giving you the solution straight away!