Claim : To show that every k-chromatic graph has a $k$-chromatic subgraph of min degree at least $k-1$.
Proof : let us assume that $G$ is $k$-chromatic graph but, it does not have a subgraph of min degree at least $k-1$, means min degree of every subgraph of $G$ is less than equal to $k-2$. This is a contradiction to our assumption that graph $G$ is $k$-chromatic (why ?).
So I am looking for the proof of the claim given above.
Hint: If you take a $k$-chromatic graph, and remove a vertex of degree $k-2$ or less, the result is still $k$-chromatic.
(Or, to put it the converse way: if you remove from a graph $G$ a vertex $v$ of degree $k-2$, and you can color $G - v$ with $k-1$ colors, then you can color $G$ with $k-1$ colors.)