Given Dedekind cuts $\alpha, \beta$, I'm trying to show that exactly one of $\alpha < \beta$, $\beta < \alpha$, or $\alpha = \beta$ show. I can prove, via considering the three successive cases in turn, that no more than one of these can occur at once, so I'm only struggling to prove that exactly one must occur at once.
My attempt is as follows. I'm going to assume that $\alpha < \beta$ and $\beta < \alpha$ fail. I'm going to show that $\alpha = \beta$ by contradiction by assuming that $\alpha \neq \beta$. Here is my attempt.
Suppose $\alpha < \beta$ and $\beta < \alpha$ do not hold. The first means that at least one of $\alpha \subset \beta$ or $\alpha \neq \beta$ fails, and the latter means that at least one of $\beta \subset \alpha$ or $\beta \neq \alpha$ fails. Suppose, for the sake of contradiction, that $\beta = \alpha$. The initial suppositions then allow us to say that $\alpha \not \subset \beta$ and $\beta \not \subset \alpha$. Therefore, there exists $x \in \alpha - \beta$ and $y \in \beta - \alpha$. But $x,y \in \mathbb{Q}$, which has a total order on it. Of course, $x \neq y$, so we either have, in $\mathbb{Q}$, $x < y$ or $y < x$. If $x < y$, then because $y \in \beta$ which is closed downward, we have $x \in \beta$, a contradiction. Similarly, if $y < x$, then because $x \in \alpha$ and $\alpha$ is closed downward, one has $y \in \alpha$, which is a contradiction. Therefore, we have $\alpha = \beta$, as required.
How does this look? I'm mainly concerned about the assumption toward a contradiction and whether I can make the assertions $\alpha \not \subset \beta$ and $\beta \not \subset \alpha$, which rely entirely on that assumption toward a contradiction.