A proof involving Dedekind Cuts

Question

If anyone could vet this proof, it would be of great assistance to me! Any feedback is welcome!

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 $x>y$

Let $D_r$ be the set $\{x \in Q, x>r\}$. Let $A$ be any dedekind cut.

We seek to prove that

1. $A \subset D_r, A \neq D_r$ if and only if there is some $q \in \mathbb{Q}-A$ such that $r<q$.
2. $A \subset D_r$ if and only if $r \in \mathbb{Q}-A$ if and only if $r<a $ for all $a \in A$.

First we prove a lemma: $c \in \mathbb{Q}-A$ if and only if $c<a$ for all $a \in A$.

Suppose $c \in \mathbb{Q}-A$. By trichotomy, either $c>a, c=a$ or $c<a$. It is obvious that $c \neq a$. If $c>a$, then by 2), $c \in A$, so it is not possible that $c>a$. Hence we are left with only $c<a$. Suppose $c>a$ for all $a \in A$. By trichotomy it is not possible that $c=a$ for any $a \in A$. Hence $c \in \mathbb{Q}-A$.

Now we prove 1).

Suppose that there is some $q \in \mathbb{Q}-A$ such that $r<q$. $q$ is thus in $D_r$ and not in $A$, and hence $D_r \neq A$. Note that $q<a$ for all $a \in A$ by the lemma. By transitivity of <, for any $ a \in A, a>r$. Hence every element of $A$ is in $D_r$. Thus, $A \neq D_r$ and $A \subset D_r$.

Now we prove the converse. Suppose $A \neq D_r$ and $A \subset D_r$. This implies that there is some element $q$ such that $q \in D_r, q \notin A$.Rephrasing this gives $q \in \mathbb{Q}-A$ and $q>r$.

Next we prove 2).

First we prove that $A \subset D_r$ if and only if $r \in \mathbb{Q}-A$. Let $r \in \mathbb{Q}-A$. Then by lemma, $r<a$ for all $a \in A$. Thus, $a \in D_r$ for all $a \in A$. This implies that $A \subset D_r$. Next we prove the converse. Let $A \subset D_r$. Then $a>r$ for all $a \in A$. From the lemma, $r \in \mathbb{Q}-A$.

We seek to prove that $r \in \mathbb{Q}-A$ if and only if $r<a$ for all $a \in A$. We note that the statement $r<a$ for all $a \in A$ is equivalent to the statement $a \in D_r$ for all $a \in A$ which is equivalent to the statement $A \subset D_r$. Hence we have already proven this by proving that $A \subset D_r$ if and only if $r \in \mathbb{Q}-A$.