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 $A \not\subseteq B$. It suffices to show $B \subseteq A$.

Since $A \not\subseteq B$, choose a rational $x$ with $x \in A$ and $x \notin B$.

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 $x \notin B$.

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 $\mathbb{R}$.