Here is the rough idea:
Suppose $C_1$ and $C_2$ overlap in at least two disjoint paths. If we follow $C_1$ along the end of one path to the beginning of the next path, and then follows $C_2$ back to the end of the first path, we obtain a cycle $C_3$. Since this cycle must have odd length, the parity of the two parts must be different. This means I can change the parity of the length of $C_1$ by following the $C_2$ part of $C_3$ instead of the $C_1$ part. This is also a contradiction, as $C_1$ has odd length.
Explicitly, let $a$ be the last vertex of one path contained in both $C_1$ and $C_2$, and let $b$ be the first vertex of the next path contained in both $C_1$ and $C_2$. Let $C_3$ be the cycle obtained by following $C_1$ from $a$ to $b$, and then following $C_2$ back to $a$. Since $C_3$ has odd length, the parity of the length along $C_1$ from $a$ to $b$ must be different from the parity of the length of $C_2$ from $a$ to $b$. It is this difference in parity that allows us to modify $C_1$. That is, let $C_1'$ be the cycle that agrees with $C_1$ except for the path from $a$ to $b$, where it agrees with $C_2$. Then $C_1'$ will have even length.
A further result has been added below:
You may want to require that distinct edges intersect in at at most one point. Without this additional assumption, even cycles can be drawn in the way you describe. (Take your A-B-C-D graph and move A and B to the edge CD, with A closer to C.)
With this extra requirement, you definitely can’t draw an even cycle.
Suppose to the contrary, and let the cycle be $v_1v_2\cdots v_{2k} v_1$. Edges $v_1v_2$ and $v_3v_4$ cross, and with the extra requirement, $v_1$, $v_2$, and $v_3$ can’t be collinear (assuming coincident vertices are also disallowed). WLOG, assume the edge $v_1v_2$ is horizontal and $v_3$ is above the line through $v_1$ and $v_2$. Then $v_4$ is necessarily below that line, $v_5$ is above, and so on. But then $v_{2k}$ and $v_3$ are on opposite sides of that line, so $v_2v_3$ and $v_{2k}v_1$ are disjoint.
Added, based on discrete’s addition to the question:
Let $G$ be a graph that can be drawn without disjoint edges. Then no vertex of $G$ can have 3 neighbors, each of which has degree at least 2.
Proof: Suppose $G$ is drawn without disjoint edges, and $v$ is adjacent to $a$, $b$, and $c$, each of degree at least $2$. We will obtain a contradiction.
[Idea of proof: The three edges $av$, $bv$, and $cv$ must all lie in one half-plane with $v$ on its boundary, and the “middle” vertex cannot have another edge that intersects the other two, as required.]
Consider the half-planes of $\mathbb{R}^2$ that are on the two sides of the line through $a$ and $v$. Except for $av$, any edge incident on $a$ or $v$ lies entirely within just one of these half-planes (the one containing its vertex that isn’t $a$ or $v$). In particular, edges $vb$ and $vc$ each lie within one of these half-planes, and so does the edge from vertex $a$ that is not the edge $av$. The latter edge must intersect both $ab$ and $ac$, so $b$ and $c$ must be in the same half-plane with respect to $av$. Similarly, $a$ and $c$ must be in the same half-plane with respect to $bv$, and $a$ and $b$ must be in the same half-plane with respect to $cv$. This is impossible. If $b$ and $c$ are in the same half-plane with respect to $av$, then either $0<\angle avb < \angle avc \le \pi$ or $0<\angle avc < \angle avb \le \pi$, and the “middle” edge of the larger angle leaves one vertex in each of the half-planes it creates.
Best Answer
Here are some hints that lead to a proof.
Show that a 2-connected graph in which every edge is on at most one odd cycle is either bipartite or an odd cycle (e.g. by using an ear decomposition).
Show that the chromatic number of a graph equals the maximum of the chromatic numbers of its blocks.
As an alternative for the second part: use induction on the number of vertices. If $G$ is 2-connected we proved it in the first part. Otherwise $G$ has a cut vertex $v$. For a component $C$ of $G-v$ show that the conditions of the problem hold in $C+v$ (since cycles must be contained completely in one such enhanced component). So you can use the induction hypothesis on each such component ($+v$) and glue them together (you can always make sure that $v$ has color 1 in each of them).