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.