Show that for any two dedekind cuts $A,B \subset \mathbb{Q}$, if $A \subset B, A\neq B$, then $B-A$ is countably infinite.

Question

Could anyone help to vet the following proof? That would be of great help to me!

A Dedekind Cut is defined as a set $D \subset \mathbb{Q}$ such that

1. $D \neq \mathbb{Q}$ and $D \neq \emptyset$
2. Let $x \in D, $ if $y \in \mathbb{Q}$ and $y > x$, then $y \in D$
3. Let $x \in D$. Then there is some $y \in D$ such that $x>y$

Lemma: Let $D$ and $C$ be any dedekind cuts such that $C \subset D$, then the set $\{x \in D \mid x<c \text{ for all }c \in C\}$ is $D-C$.

Proof of Lemma:

Let $x \in D$ and $x \notin C$. Let $y \in C$. By trichotomy, either $x<y, x=y$ or $x>y$. It is obvious that $x \neq y$, because otherwise $x \in C$.By 2) in the definition of dedekind cuts, it is also not possible that $x>y$. Hence $x<y$ must be true.

Proof that $B-A$ is countably infinite:

Since $B-A \subset \mathbb{Q}$ and $\mathbb{Q}$ is countably infinite, $B-A$ will be a countable set. Hence $B-A$ will either be countably infinite or finite. We merely need to show that for $A-B$ is not a finite set. Suppose that $B-A$ is a finite set. We can be assured that $B-A$ is not empty, since $B \neq A$. So $B-A=\{x \in B \mid x<a \text{ for all }a \in A\}=\{x_1,x_2...x_n\}$ where $n \in \mathbb{N}$. Let $z=$ min$\{x_1,x_2,x_3...\}$. Since $z \in B,$ there exists $w$ such that $w<z, w \in B$. By transitivity of <, $w<a$ for all $a \in A$, and hence $w \in B-A$ . But this is not possible, since $w \notin \{x_1,x_2,x_3...x_n\}$. Thus, $B-A$ cannot be finite, since we can always find such an element $w$. It follows that $B-A$ can only be a countably infinite set.