Average degree of $k$-degenerate graph is $\leq 2k$ How to prove the following claim?

Average degree of $k$-degenerate graph is $\leq 2k$

Definition: Graph is $k$-degenerate if for every $\,G' = (V',E') \subset G$ there exists $v 
\in V'$ such that $\deg(v) \leq k$.
Formalization of the statement: $\sum_{v \in V}{\deg(v)}/ \lvert V \rvert \leq 2k$.
It seems to me there is some knowledge about $k$-degenerate graphs I am missing.
 A: By coincidence this was proved yesterday on
Girth and monochromatic copy of trees
The first part of the answer shows that every graph $G=(V,E)$ has a subgraph with minimum degree at least $\frac{|E|}{|V|}$, which is half of the average degree.
I trust you can take it from there.
(ADDED)
Assume the average degree of our $k$-degenerate graph $G$ is larger than $2k$.
Then $G$ has a subgraph with minimum degree larger than $k$.
This contradicts $k$-degeneracy.
A: Take graph G, as it's k-degenerate, there is a vertex of at most degree k, call it v. Removing v reduces the number of the edges by k and therefore reducing the sum of the degrees by at most 2*k.
Now, take the vertex u with the minimum degree of G / {v} and remove it; degree(u)<= k in G / {V}  as G is k-degenerate, therefore you will remove at most k edges, i.e. the sum of the degrees of the remaining graphs will be reduced by at most 2k. If you repeat this procedure for n-1 times, the sum of the degrees will be reduced by 2k * (n-1) and essentially this will be an upper bound on the sum of the degrees of G. Hence, the average degree would be 2k * (n-1)/n <= 2k.
