If a graph has no cycles of odd length, then it is bipartite: is my proof correct?

I came up with a proof of

Graph $G$ has no cycles of odd length $\implies$ $G$ is bipartite.

like this:

Without loss of generality, let's only consider a connected component, because if every connected component of a graph is bipartite, then the whole graph is bipartite.

Pick up a random vertex $v$ in $G$, calculate the length of the shortest simple path from $v$ to any other node, call this value distance from $v$, and divide nodes into 2 groups according to the parity of their distance to $v$. If we can prove that nodes belong to the same group can not be adjacent, then we know that we actually get a partition of the $G$ that fulfill the definition of bipartite graph.

Now, to introduce contradiction, assume two nodes $x, y$ with both even or odd distance from $v$ are adjacent, then the shortest simple path $\langle v, x \rangle$, $\langle v, y \rangle$ and edge $\{x, y\}$ contains a cycle with odd length, which is contradictory to that $G$ has no cycles of odd length. In other words, nodes both with even or odd distance from $v$ can not be adjacent, which is exactly what we need.

So my question is, is my proof correct? And is there simpler method to prove the proposition?

Edit:

(to address comment from Srivatsan Narayanan)

To prove that $\langle v, x \rangle$ and $\langle v, y \rangle$, together with $\langle x, y \rangle$ contains a cycle with odd length is obvious when $\langle v, x \rangle$ and $\langle v, y\rangle$ are disjoint. When that's not the case, let's give the last node shared by $\langle v, x\rangle$ and $\langle v, y\rangle$ the name $v'$. So the three nodes $v', x, y$ forms a cycle with length $\newcommand{\len}{\operatorname{len}}$

$$ L = \len(\langle v', x \rangle) + \len(\langle v', y \rangle) + 1 = \len(\langle v, x \rangle) + \len( \langle v, y \rangle) - 2 \cdot \len(\langle v, v'\rangle) + 1 .$$

where $\len()$ means the length of the shortest path.

As $\len(\langle v, x \rangle)$ and $\len(\langle v , y \rangle)$ are both even or odd, then $L$ must be odd. Therefore, in both cases, disjoint or not, $\langle v, x \rangle$, $\langle v, y \rangle$ and $\langle x, y\rangle$ contains a cycle with odd length.

Edit2

To see $\len(\langle v, x \rangle) = \len(\langle v, v'\rangle) + \len(\langle v', x \rangle)$, we can simply prove that both $\langle v, v' \rangle$ and $\langle v', x \rangle$ are both shortest path. And that's obvious, because if it's not the case, there exist a path shorter than $\langle v, v' \rangle$ from $v$ to $v'$, or there exist a path shorter than $\langle v', x\rangle$ from $v'$ to $x$, then $\langle v, x \rangle$ can not be a shortest path.

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.