Let $G=(V,E)$ be a graph with minimum degree $\delta(G)=n\lt\aleph_0$. Does $G$ necessarily have a spanning subgraph $G'=(V,E')$ which also has minimum degree $\delta(G')=n$ and is minimal with that property?
This question is easily answered in the affirmative if $G$ is locally finite or if $n\le1$. It already seems difficult for $n=2$, but I am not very clever and may be missing something obvious.
The question also seems to make sense for hypergraphs:
Let $m,n\in\mathbb N$. Let $E$ be a family of sets, each of cardinality at most $m$. If $E$ is an $n$-cover of a set $V$ (each element of $V$ is in at least $n$ elements of $E$), does $E$ contain a minimal $n$-cover of $V$?
I would expect such simple questions to have been asked and answered 100 years ago.
Where are these questions considered in the literature?
P.S. The following proof for the simple case of a (non-hyper) graph with $\delta=1$ is probably a dead end, as it does not seem to generalize in any obvious way. I'm putting it here anyway because it's quite simple.
Theorem. A graph with no isolated points has a minimal spanning subgraph with no isolated points.
Proof. Let $G$ be a graph with no isolated points. Let $H$ be a maximal spanning subgraph of $G$ not containing $K_3$ or $P_4$ as a subgraph, induced or otherwise. Then $H$ is a star-forest, possibly with some isolated points. For each isolated vertex $v$ of $H$, choose an edge of $G$ which is incident with $v$ and add it to $H$. This results in a spanning subgraph of $G$ in which each component is a nontrivial tree of radius at most $2$.The proof is completed by observing that any nontrivial tree of radius at most $2$ has a spanning subgraph with no isolated points.