Let $E, F \in \mathsf{Min}$ such that $E \subseteq F$. Denote $n : = \min(E)$ and $m : = \min(F)$. Now $n \preceq x$ for every $x \in E$ and $m \preceq y$ for every $y \in F$. Since $E$ is contained in $F$, then particularly $m \preceq x$ for every $x \in E$. Since $n \in E$, this implies that $m \preceq n$. $\square$