Okay, so:
Firsly, let's take any $q\in Q\cap A$. We know that $ q \in A$, so there exists $\epsilon_q$, such: $(q-\epsilon_q;q+\epsilon_q) \subseteq A$.
Now, we're ready to look at $\alpha(q),\beta(q)$. Why do they exist? A is bounded, $[q-\frac{\epsilon_q}{2}; q] \subseteq A $, and $[q;q + \frac{\epsilon_q}{2}] \subseteq A$.
It is clear now, that both $\alpha(q),\beta(q)$ exists and $\alpha(q) < \beta(q)$.
I suppose u denote $I_q = (\alpha(q),\beta(q)) $.
$I_q \subseteq A$ because $\alpha(q) \ (\beta(q))$ by definition are such that $(\alpha(q),q] \subseteq A$ and $[q,\beta(q)) \subseteq A \ \ \ \ (*)$
To prove 1 and 2:
1)
I'll show two inclusions. Let's pick any $x \in \bigcup_{q\in Q \cap A} I_q $. I'll show that $x \in A$. Clearly for particular $q_x$, $x \in I_{q_x}$, that means $ x \in (\alpha(q_x),\beta(q_x)) $ and by $(*)$ we get $x \in A$
Now let's pick any $y \in A$. There exists $ \epsilon_y $ such that $(y-\epsilon_y;y+\epsilon_y) \subseteq A $. In any interval, there is $ p \in Q $, $p \in (y-\epsilon_y;y+\epsilon_y), p>y $. Let's look at $I_p = (\alpha(p),\beta(p)) $.
I'll show that $\alpha(p)<y$, which is obvious because by definition of $\alpha(p)$, there must holds inequality: $\alpha(p) \leq y-\epsilon_y < y $.
So $ y \in (\alpha(p),p) \subseteq I_p \in \bigcup_{q \in Q \cap A} I_q $
2)
If $I_q = I_s$ everything's good, if no, let's assume that $I_q \cap I_s \neq \emptyset$
Just to fix something, say $q<s$ ($s<q$ case is really the same).
Clearly then $\beta(q) < \beta(s) $, but $\beta(q) \in (\alpha(s),\beta(s))$ that means $\beta(q)$ cannot be $ \sup\{x\in R: [q,x) \in A\} $ (because $[q,\beta(q)) \in A , (\alpha(s), \beta(s)) \in A , \alpha(s) < \beta(q) < \beta(s)$ (otherwise it really would be $I_q \cap I_s = \emptyset $)
So,there we have a problem, cause whole $[q,\beta(s)) \subseteq A$, and that means $\beta(q)$ wasn't supremum. Contrary