TITLE: Directing graph such that any outdegree would be at most 2
QUESTION [6 upvotes]: Let $G=(V,E)$ be a graph. Suppose that for every subgraph $G'=(V',E') , G' \subset G$, we have $|E'| \le 2|V'|$. Show it's possible to direct $G$ such that any outdegree would be at most $2$.

I tried proving it by induction. I've realized that the minimum degree at any subgraph of $G$ is at most 4. I tried removing the vertex $v_0$ with the minimal degree at a certain subgraph from the original graph, claiming that the graph from the remaining vertexes can be directed as required. How can I include the vertex $v_0$ in the new directed graph such that any outdegree would still be at most 2?
Thanks

REPLY [5 votes]: We proceed via induction, over the number of edges.
Take a graph $G$ and remove an edge $e$, suppose its vertices are $u$ and $v$. By the inductive hypothesis $G- e$ can be directed so that all out-degrees are $2$ or less. If one of $u$ and $v$ has out-degree less than $2$ we are done.
Otherwise consider the set $R$ of all vertices reachable from $u$ or $v$ by using the directed edges. If there is a vertex $x\in R$ with out-degree $1$ or $0$ we are done, simply take the path (that goes from $u$ or $v$ to $x$) and reverse every edge, this leaves one of $u$ or $v$ with outdegree $1$.
Suppose no vertex in $R$ has out-degree less than $2$. Then the number of edges in $G-e$ between vertices in $R$ is $2|R|$, and thus adding edge $e$ we get $2|R|+1$ edges, a contradiction.