Inclusion totally orders the Dedekind reals

The relation $\le$ is set inclusion, and $\subseteq$ is reflexive, antisymmetric (mutual inclusion $A\subseteq B$ and $B\subseteq A$ forces the set equality $A=B$), and transitive; hence $\le$ is a partial order on $\mathbb{R}$.

It remains to establish totality. Fix cuts $A,B$; if $A\subseteq B$ there is nothing to prove, so assume $A\subseteq ̸B$. It suffices to show $B\subseteq A$.

Since $A\subseteq ̸B$, choose a rational $x$ with $x\in A$ and $x\mathrm{\notin }B$.

Every $y\in B$ satisfies $y<x$: otherwise $x\le y$ by trichotomy, and then downward closure of $B$ places $x\in B$ (directly if $x<y$, or as $x=y\in B$), contradicting $x\mathrm{\notin }B$.

Fix any $y\in B$. From $y<x$ together with $x\in A$, downward closure of $A$ gives $y\in A$; as $y\in B$ was arbitrary, $B\subseteq A$.

Thus for all cuts $A,B$, $A\subseteq B$ or $B\subseteq A$, so $\le$ is total; combined with the partial-order properties, set inclusion is a total order on $\mathbb{R}$.