Given is a connected graph with undirected edges such that deleting any vertex decomposes the graph into $\le k$ components, for some $k\ge 2$. Is it true that the graph has a spanning tree with degree at most $k$? 


The claim is clearly true if the graph itself is a tree. I checked some articles on vertex connectivity but couldn't find anything related to this. 


**Edit**: Tony Huynh found a counterexample. What if the graph is required to be such that removing some vertex does not give a connected graph?



Deleting vertex decomposes graph
No, this is false. Take $K\_{n, n+2}$. Removing any vertex gives a connected graph, so in particular the graph satisfies the condition with $k=2$. However, it does not contain a spanning tree with maximum degree $2$, since $K\_{n, n+2}$ does not contain a Hamiltonian path.


Here is a counterexample that works for all $k$. Let $G\_n$ be the graph obtained from the graph consisting of $n$ parallel edges by subdividing each edge once. This graph is $2$-connected and in particular satisfies the condition for $k=2$. However, every spanning tree of $G\_n$ has maximum degree at least $\frac{n}{2}$. 


