Let $H=(V,E)$ be a hypergraph, that is $V$ is a set and $E\subseteq \mathcal{P}(V)$. We say that $C\subseteq E$ is a *cover* of $H$ if $\bigcup C = V$.


A cover $M\subseteq E$ is said to be *strongly minimal* if for every cover $C$ of $H$ we have $$\text{card}(M\setminus C)\leq \text{card}(C\setminus M).$$


Is there a hypergraph $H=(V,E)$ with the following properties?


1. $\bigcup E = V$;
2. $e\in E$ and $e'\subseteq e$ implies $e'\in E$;
3. for all $e\in E$ there is $m\in E$ such that $e\subseteq m$ and $m$ is maximal in $E$ with respect to set inclusion;
4. $H$ does not have a strongly minimal cover.



Strongly minimal covers
Maybe I misunderstood something, but consider the following simple example.
Let $V=(0,\infty)$ and let the maximal edges be the (open) unit intervals, except $(0,1)$.
Any cover contains a sequence converging to $0$, so in fact there isn't any minimal cover at all.


**Update:** Noah asked whether there is an example where all edges are finite.
Here I give such an example.
Take $V=\{\frac 1i\mid i\in \mathbb N\} \cup \{-\frac 1i\mid i\in \mathbb N\}$.
For any $n\in \mathbb N$ we define two maximal edges.
The first is $\{\frac 1i\mid i\le n\} \cup \{-\frac 1n\}$ and the second is $\{-\frac 1i\mid i\le n\} \cup \{\frac 1n\}$ (note that these are the same for $n=1$, but this won't matter).
Again there isn't any minimal cover.


