TITLE: Every graph $G$ contains a subgraph $H$ such that $\deg_H(v) \geq \chi(g) -1$ for every vertex
QUESTION [5 upvotes]: I stumbled upon following theorem for which no proof was given:

Let $G$ be an undirected and simple graph. Then there is a subgraph $H$ of $G$ such that $\deg _H (v) \geq \chi(G) - 1$ for every vertex $v \in V(H)$.

I came up with following proof:
Proof. Assume the statement fails for some graph $G$. We first observe that $G$ cannot satisfy $\chi(G) < 2$ since otherwise one could choose an arbitrary vertex for $H$. We will now prove the following claim:
Claim. Every subgraph $H$ of $G$ is $(\chi(G) - 1)$-colorable. 
Proof of the claim. By induction on $n = |V(H)|$. If $n = 1$ the claim follows by our observation that $\chi(G) \geq 2$. For arbitrary $n$ we choose $v \in V(H)$ such that $\deg_H(v) < \chi(G) - 1$. Now, $H - v$ is $\chi(G) -1$ colorable by the induction hypothesis. Therefore, $H$ is also $(\chi(G)-1)$-colorable by our choice of $v$. This yields the claim.
Continuing our proof, $G$ is also a subgraph of itself (we never made use of the fact that $H$ was a proper subgraph). Hence $G$ is $(\chi(G) - 1)$-colorable which yields a contradiction. q.e.d.

Can anyone verify the correctness of this proof? I am not 100% certain that it is actually correct. Are there simpler ways to prove the theorem?

REPLY [3 votes]: The proof is fine.
To put this theorem in context: we say that a graph $G$ is $k$-degenerate if every subgraph of $G$ has a vertex of degree at most $k$. One of the basic result about these graphs is that, whenever $G$ is $k$-degenerate, $\chi(G)$ is at most $k+1$.
This is shown by a similar argument to yours. If $G$ is $k$-degenerate, then $G$ has a vertex $v$ of degree at most $k$. Every subgraph of $G-v$ is a subgraph of $G$; therefore $G-v$ is $k$-degenerate, so by induction we can $(k+1)$-color it. We can extend this to a $(k+1)$-coloring of $G$ by giving $v$ a color not present on any of its (at most $k$) neighbors.
The contrapositive of this result is that if $\chi(G) = k$, $G$ cannot be $(k-2)$-degenerate. In other words, not every subgraph of $G$ has a vertex of degree at most $k-2$; in some subgraph of $G$, every vertex has degree at least $k-1$.