Split graph in two parts, such that most nodes have even number of edges
By induction , we can prove that a partition exists with all nodes having even degree within the partition.
If all vertices have even degree , then we are done (Just partition it into the whole graph and an empty graph)
Otherwise let $v$ be a node with odd degree , now modify the graph by inverting the neighbourhood of $v$ (If $a$ and $b$ are neighbours of $v$ , then if they have an edge , remove it otherwise add it) and also remove $v $ from the graph , for the new graph , split it into 2 parts $A$ and $B$ such that all nodes in $A$ and $B$ have even degree , add $v$ back to $A$ if $A$ has even neighbours of $v$ , else add it to $B$. This works because in $A$(Or $B$ if we added $v$ to $B$) all degrees are even except for neighbours of $v$ which become even after adding $v$