Splitting up the nodes of a Graph such that all nodes in these graphs have even degree. $G$ is a simple undirected graph. Show that
2 disjunctive sets $A$ and $B$ exist such that $A \cup B = V(G)$ and all nodes in $G[A]$ and $G[B]$ have even degree.
I have already proven that this is true for trees and cycles.
Sadly I have to idea generalize that.
It would be really great if someone could give me a hint.
 A: Can be proved by induction on number of vertices:
Base case: if $G$ is a graph on less than 3 vertices, then the claim is true. for example, if the graph is $k_3$ then $A=V(G),B=\{\}$ is a desired partition.
Suppose the claim is true for all graph on $n$ vertices and we are given a graph $G$ on $n+1$ vertices, then  if all vertices of $G$ is even then done, as we can take $A=V(G)$ and $B=\{\}$. So suppose there is a vertex $v\in V(G)$ of odd degree, then construct a new graph $G'$ from $G$ by removing the vertex $v$ and change the induced subgraph of the resulting graph induced by the vertices $N(v)$ by its complement graph. Then by induction hypotheses, there is a partition $(A,B)$ of $G'$ such that the vertices of the induced graph on A and B all have even degree. Now one of $A \cap N(v)$ or $B \cap N(v)$ is even as $|N(v)|$ is odd. So Suppose W.L.O.G, that $|A \cap N(v)|$ is even, then consider partition $A'=A\cup \{u\}, B'=B$ of $G$. We have that vertices $u \neq N(v) \cup{v} $ have even degree in induced subgraph on $A'$ or $B' $. $v$ has even degree in the induced subgraph on $A$ by construction of $A$. The parity of degree of a vertex $u\in N(v) \cap A$ is flipped twice going from a vertex in induced graph on $A$to $A'$ once when we added $v$ to $A$, once we change the the subgraph spanned by $N(v)$ to its complement. The parity of degree of a vertex in $B \cap N(S)$ is unchanged as there are odd number of vertices in $B \cap N(S)$ and so a vertex  there has at most  even number of neighbour(Think about this,  I have hard time explaining) so when you look at complement graph of line graph induced by $B\cap N(S)$, the parity of a vertex stays the same.  So  that' s it.
