If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$: Proving Two Implications. Principles of Mathematical Analysis By Walter Rudin I am told that the following proposition (II) is true by definition. I then attempt to prove the two implications that follow from (II) being true.

The members of $R$ will be certain subsets of $Q$, called cuts. A cut is, by definition, any set $\alpha \subset Q$ with the following three properties.
(I) ...
(II) If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$.
(III) ...
The letters $p, q, r, ...$ denote rational numbers, and $\alpha, \beta, \gamma, ...$ will denote cuts.
(II) implies ...
If $p \in \alpha$ and $q \not\in \alpha$, then $p < q$.
If $r \not\in \alpha$ and $r < s$, then $s \not\in \alpha$.

Proof of Theorem 1.19, Step 1, Principles of Mathematical Analysis By Walter Rudin.

First Proof
Let $p \in \alpha$ and $q \not\in \alpha$.
$\alpha = \{ p,q \in Q : q < p \}$ From (II).
$q \not\in \alpha \rightarrow q \in \{ p,q \in Q : q > p \}$ Since an object is a member of a set if and only if it satisfies the defining property of the set.
$Q.E.D.$
Second Proof
Let $r \not\in \alpha$ and $r < s$.
$\alpha = \{ p,q \in Q : q < p \}$ From (II).
Since we had $r \not\in \alpha$ in the hypothesis, we can say that there is no $q \in Q$ such that $q < r$.
Therefore, $q > r \ \forall q \in Q, r \not\in \alpha$
Let $s = q$.
Since we have $r \not\in \alpha$, $s \in \{ s, r \in Q : s > r \}$
Therefore, we can conclude that $s \not\in \alpha = \{ s, r \in Q : s < r \}$
$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proofs for correctness.

New Second Proof
Let $r \not\in \alpha$ and $r < s$.
$\{ q \in Q : q < p \} \subseteq \alpha$ This is from (II).
$r \not\in \alpha \rightarrow r \not\in \{ q \in Q : q < p \} \subseteq \alpha$
$r \in \{ q \in Q : q > p \}$ Since if $q = p$, then $q = r \in \alpha$.
Since we have that $r < s$, $r \in \{ q \in Q : s > q > p \}$
But how can I conclude from here that $s \not\in \alpha$?
 A: Well, in both cases, you should note that your definition of cut $\alpha$ is not proper since:
$$\alpha=\{p,q\in \mathbb{Q}:q<p\}$$
is not clear enough. By that, I mean that, when describing a set, at left you note the elements of this set of which "greater" set are elements, hence, when you say that $\alpha$ consists of $p,q\in\mathbb{Q}$, it is not clear whether you intend that both $p,q$ belong to $\alpha$ or not. It would be more proper to mention that:
$$\{q\in\mathbb{Q}:q<p\}\subseteq\alpha\tag{rule}$$
Now, we can use the following rule from mathematical logic:
$$(A\Rightarrow B)\Leftrightarrow(\neg B\Rightarrow\neg A)$$
Now, for your first proof, let $p\in\alpha$ and $q\in\mathbb{Q}$. We know that:
$$q<p\Rightarrow q\in\alpha$$
From our rule, we have that:
$$(q<p\Rightarrow q\in\alpha)\Leftrightarrow(q\not\in\alpha\Rightarrow q\geq p)$$
Now, since $p\in\alpha$ and $q\not\in\alpha$, it is clear that $p\neq q$, so, we have that $q>p$.
Can you, now, use these thoughts and your own ideas to alter you second proof?
Edit: As far as why $\alpha_p:=\{q\in\mathbb{Q}:q<p\}\subseteq\alpha$ and not neccessarily $\alpha_p=\alpha$, it is not guaranteed by property $(II)$ that $\alpha$ even represents a rational number, so no-one can ensure us that $p$ is the $\max\alpha$ or the $\sup\alpha$. So, there might exist $s>p$ such that $s\in\alpha$, which means that $\alpha_p\subset\alpha$.
Further edit: As for the second proof, note at first that, for two sets $A,B$ if $A\subseteq B$ does not mean that $B\subseteq A^c$, where with $A^c$ we note the complementary set of $A$. So, it is not true that:
$$\alpha\subset\{q\in\mathbb{Q}:q\geq p\}$$
Moreover, as for $s$, it cannot belong in the set $\{q\in\mathbb{Q}:s>q>p\}$ (can you see why?). Try to mimic the proof of the first claim using the first claim and the rule described.
As a further explanation to the rule, what it is stated is that, you can say that "if $A$ is true, then $B$ is true" if and only if you can say that "if $B$ is not true, then $A$ is not true".
A: Note that $\alpha$ is not given explicitly as a set, it's given that it's a set with specific properties. Your approach to describe $\alpha$ as a set using $p, q$ is wrong. The implications can be proved easily using contradiction (assume hypotheses and negative of conclusion).
Thus to prove the first implication assume that $p\in\alpha, q\notin\alpha$ (hypotheses) as well as $q\leq p$ (negative of the conclusion). Now if $q=p$ then $q\in\alpha$ because $p\in\alpha$. And if $q<p$ then by the property (II) $q\in\alpha$. This is the contradiction which proves the implication. You can try to prove the other implication as well in a similar manner.
Moreover you should try to go beyond the symbols and think about their meaning in your natural language. Thus if $p\in\alpha$ then by second property we see that all numbers smaller than $p$ are also in $\alpha$. So whatever is not lying in $\alpha$ must necessarily be greater than $p$. 

You may ask why we can't describe the set $\alpha$ in the manner you are trying. Well, the properties of $\alpha$ are meant to describe a general cut and not any specific chosen cut. All the cuts satisfy those properties and using those properties alone you can't describe a specific cut.
