Let $H=(V,E)$ be a hypergraph, $\kappa$ a cardinal.

Observation 1. $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ if and only if $H$ has a maximal $\kappa$-colorable subgraph.

Proof. If $F\subseteq E$, then $(V,F)$ is a maximal $\kappa$-colorable subgraph of $H$ iff $E\setminus F$ is a minimal element of $\mathcal M(H,\kappa)$.

Observation 2. If $\kappa$ is infinite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ iff $H$ is $\kappa$-colorable. (Hence we get a simple example where $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$ by taking $\kappa=\aleph_0$ and $H=K_{\aleph_1}$, the complete graph on $\aleph_1$ vertices.)

Proof. For the nontrivial direction, since $\kappa+1=\kappa$, we can always make another set non-monochromatic by creating a new color.

Observation 3. If $\kappa$ is finite and all hyperedges (elements of $E$) are finite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$.

Proof. By Observation 1, we have to prove that $H$ has a maximal $\kappa$-colorable subgraph. This follows from Zorn's lemma and the hypergraph version of the De Bruijn–Erdős theorem, which says that for finite $\kappa$ a hypergraph with finite hyperedges is $\kappa$-colorable iff all of its finite subhypergraphs are $\kappa$-colorable.

Observation 4. If $1\lt\kappa\lt\aleph_0$ and if $E$ is a nonprincipal ultrafilter on $V$, then $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$.

Proof. Consider any coloring $c:V\to\kappa$. Since $E$ is an ultrafilter and $\kappa$ is finite, there is some $i\in\kappa$ such that $V_i=\{v\in V:c(v)=i\}\in E$, whence $V_i\in\operatorname{Mono}(H,c)$. Now we can make $V_i$ non-monochromatic by changing the color of one vertex in $V_i$; and this will not result in any previously non-monochromatic hyperedge becoming monochromatic, because each hyperedge has infinite intersection with $V_i$. Therefore $\operatorname{Mono}(H,c)$ is not a minimal element of $\mathcal M(H,\kappa)$.

Let $H=(V,E)$ be a hypergraph, $\kappa$ a cardinal.

Observation 1. $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ if and only if $H$ has a maximal $\kappa$-colorable subhypergraph.

Proof. If $F\subseteq E$, then $(V,F)$ is a maximal $\kappa$-colorable subhypergraph of $H$ if and only if $E\setminus F$ is a minimal element of $\mathcal M(H,\kappa)$.

Observation 2. If $\kappa$ is infinite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$ if and only iff $H$ is $\kappa$-colorable. (Hence we get a simple example where $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$ by taking $\kappa=\aleph_0$ and $H=K_{\aleph_1}$, the complete graph on $\aleph_1$ vertices.)

Proof. For the nontrivial direction, since $\kappa+1=\kappa$, we can always make another hyperedge non-monochromatic by creating a new color.

Observation 3. If $\kappa$ is finite and all hyperedges (elements of $E$) are finite, then $\operatorname{Min}(\mathcal M(H,\kappa))\ne\emptyset$.

Proof. By Observation 1, we have to prove that $H$ has a maximal $\kappa$-colorable subhypergraph. This follows from Zorn's lemma and the hypergraph version of the De Bruijn–Erdős theorem, which says that for finite $\kappa$ a hypergraph with finite hyperedges is $\kappa$-colorable iff all of its finite subhypergraphs are $\kappa$-colorable.

Observation 4. If $1\lt\kappa\lt\aleph_0$ and if $E$ is a nonprincipal ultrafilter on $V$, then $\operatorname{Min}(\mathcal M(H,\kappa))=\emptyset$.

Proof. Consider any coloring $c:V\to\kappa$. Since $E$ is an ultrafilter and $\kappa$ is finite, there is some $i\in\kappa$ such that $V_i=\{v\in V:c(v)=i\}\in E$, whence $V_i\in\operatorname{Mono}(H,c)$. Now we can make $V_i$ non-monochromatic by changing the color of one vertex in $V_i$; and this will not result in any previously non-monochromatic hyperedge becoming monochromatic, because each hyperedge has infinite intersection with $V_i$. Therefore $\operatorname{Mono}(H,c)$ is not a minimal element of $\mathcal M(H,\kappa)$.