Suppose you have a graph on vertices $v_1, v_2, v_3, …, v_n$.
Now, an operation starts.
Step 1: remove all edges $v_1v_k$ for such k that $deg(v_k) \ge deg(v_1)$
Step 2: remove all edges $v_2 v_k$ for such k that $deg(v_k) \ge deg(v_2)$ (here, $deg(v_i)$ denotes degree of $v_i$ in the graph obtained on step 1, not in the original graph. Also, $v_2v_1=v_1v_2$)
And so on until step n.
Question: Is it true that after such procedure the remaining graph will contain an isolated vertex?
I have no idea how to prove or disprove this, as the procedure is quite complicated. It is obviously true for trees (as they contain a vertex of degree 1) and complete graphs.
False proof: if $deg(v_i) \le deg(v_j)$ in the original graph, the edge $v_iv_j$ will be removed on step i, so all vertices will be isolated.
Why it is false: before we arrive to step i, $deg(v_i)$ and $deg(v_j)$ may change, so the edge may remain.
The idea of using the vertex of minimal degree fails for the same reason: when we consider step corresponding to that vertex, it can already be non-minimal (as a consequence of previous steps)
Any hints are welcome.
By the way, here's a code for Wolfram Mathematica to check this for a random graph with 10 vertices and 40 edges (I didn't find counterexamples, but may be the code is flawed?) [Be wary, if you want to change 10 to something else, it should be done in three places (in the beginning and in the cycles' conditions )] :
g = RandomGraph[{10, 40}, VertexLabels -> Placed[Automatic, Center], VertexSize -> .5];
gg = g
For[i = 1, i < 11, i++, For[j = 1, j < 11, j++, If[i != j && MemberQ[EdgeList[g], Min[i, j] \[UndirectedEdge] Max[j, i]] && VertexDegree[g, j] >= VertexDegree[g, i], g = EdgeDelete[g, Min[i, j] \[UndirectedEdge] Max[j, i]], ]]]
g
Best Answer
A quick run with Sage found easily some counter-examples for your conjecture, on 10 vertices.
It has degree sequence $[7, 6, 6, 6, 5, 5, 4, 4, 4, 3]$. After applying your algorithm, we end up with
Here is my code for information
Here is the full details of the operations,step by step.