In the article Forcing highly connected subgraphs from Maya Jakobine Stein there is a statement about a theorem of Mader which guarantees the existence of highly connected subgraphs of a given graph with a high minimum degree. Then a well known theorem of Mader which guarantees the existence given a high average degree is stated.
Theorem 1. Any finite graph $G$ of average degree at least $4k$ has
a $(k+1)-$connected subgraph.
A result of the article is the following:
Theorem 2. Let $k\in\bN$ and let $G$ be a graph such that each vertex has degree
at least $2k$ and each end has edge-degree at least $2k$. Then $G$ has a $(k + 1)-$edge connected
region.
Theorem 1 doesn't guarantee by itself the existence of such a subgraph only given a minimum degree of $2k$ even in the finite case. The source of the theorem, which, unfortunately for me, is in German, may have such a result, but fortunately a simple proof of the following proposition follows from the steps used by Stein to prove Theorem 2.
Proposition 1. Let $k\in\bN$ and let $G$ be a finite graph with minimum degree $\delta(G)=2k$. Then $G$ has a $(k + 1)-$edge connected subgraph.
To understand a bit of Stein's proof, we have to know some concepts, like ray, end, region, boundary, and to adapt the proof to our finite case we must make sure none of those concepts are relevant for the finite case.
So, given a graph $G$ and a subgraph $H$, the boundary of $H$ is just the set $N(G-H)$ of all vertices of $G$ adjacent to some vertex in $G-H$. A region is a connected induced subgraph with finite boundary. An infinite path starting at some vertex is called a ray, and the ends are the equivalence classes of rays given a specific equivalence relation for them. The edge-degree of an end is the maximum cardinality of a set of edge-disjoint rays in it. So, indeed, since a finite graph has no rays, then it has no ends, and the corresponding hypothesis becomes a void one for the finite case. Also in a finite graph the boundary of any subgraph is obviously finite, so a region in a finite graph is just an induced subgraph.
With this we can say that the same proof should work in the finite case, but we also should be able to make it into a simpler proof. First, the proof is based on the following lemma:
Lemma. Let $D\neq \varnothing$ be a region of a graph $G$ so that $|E(D, G − D)| < m$ and so that $|E(D' , G − D')| ≥ m$ for every non-empty region $D' \subseteq D − \partial D$ of $G$. Then there is an inclusion-minimal region $H \subseteq D$ with $|E(H, G − H)| < m$ and $H\neq \varnothing$.
The lemma is used to find a minimal region $C$ such that $|E(C, G − C)| < 2k$. But this lemma is just a big nuke for our finite case: If $G$ is not $(k+1)-$edge connected such a set $C$ exists: Take a minimum separating set $S$ of $G$ with $k$ edges, and a component $H$ of $G-S$. Such a component is a choice for the set $C$, and a good one since $|E(H,G-H)\leq k<2k$. The existence of a minimal one is also immediate: Any finite poset has a minimal element. Even more, we can take one of minimum size. So, for the following proof, which is in fact the last paragraph of the proof in the article (with a few more details), this set will be called $H$.
Proof of Proposition 1. Let $H$ be as in the comments above. We claim that $H$ is $(k+1)-$edge connected. Indeed, if it were not, then $|H|\geq 2$ (If $|H|=1$ then $V(H)=\{v\}$ for some $v\in V(G)$ and since $d(v)\geq 2k$, then $|E(H,G-H)|=d(v)\geq 2k$). Furthermore, $H$ has a minimum separating set $K'$ with $|K'|\leq k$. Then $H-K'$ has at least two components, say, $H',H''\subsetneq H$. If $$|E(H,G-H)\cap E(H',G'-H')|> \frac{E(H,G-H)}{2}$$
then $$|E(H'',G-H'')\cap E(H,G-H)|\leq \frac{E(H,G-H)}{2}< k.$$
So, given that, the following holds
$$E(H'',G-H'')\leq |E(H'',G-H'')\cap E(H,G-H)|+|K'|<2k$$
otherwise, if $|E(H,G-H)\cap E(H',G'-H')|\leq \frac{E(H,G-H)}{2}<2k$
then
$$E(H',G-H')\leq |E(H',G-H')\cap E(H,G-H)|+|K'|<2k.$$
Both cases contradict the minimality of $H$. Therefore, $H$ is $(k+1)-$edge connected.
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