Theorem 12.3 from Matsumura Theorem 12.3 (p. 87), Commutative Ring Theory by Matsumura.

Let $A$ be a Krull ring, $K$ its field of fractions, and $\mathfrak{p}$ a height $1$ prime ideal of $A$; then if $\mathcal{F} = \{R_{\lambda}\}_{\lambda \in \Lambda}$ is a family of DVRs of $K$ defining $A$, we must have $A_\mathfrak{p} \in \mathcal{F}$. If we set $\mathcal{F}_0 = \{A_\mathfrak{p} | \mathfrak{p} \in \text{Spec} \space A \space \text{and} \space \text{ht} \space \mathfrak{p} =1\}$ then $\mathcal{F}_0$ i sa defining family of $A$. Thus $\mathcal{F}_0$ is the minimal defining family of DVRs of $A$.

I got stuck in two places while reading this proof. I will provide the proof below in case anyone wants to see it. 
Question 1
I'm not really sure why the following two statements are equivalent:
1) for $a, b \in A$ with $a \not= 0 $, $b \in aA_\mathfrak{p}$ for all $A_\mathfrak{p} \in \mathcal{F_0} \implies b \in aA$
2) $aA$ can be written as the intersection of height $1$ primary ideals. 
Question 2
Suppose that for $x \in K$, there exists $ 0 \not=a \in A$ such that $ax^n \in A$ for all $n>0$. Then, if $A$ is Noetherian, $x$ must be integral over $A$. (I want to prove this to understand the very last part where it says that $A$ is completely integrally closed). 
Proof From the Textbook




 A: It's easily seen that $2)\Rightarrow 1)$ (and this is all that matters for the proof): Suppose that $aA$ is a finite intersection of primary ideals whose radicals are height one primes. For such a prime $p$, from $b\in aA_p$ we get an $s_p\in A-p$ such that $s_pb\in aA$. In particular, $s_pb$ belongs to the primary ideal in the primary decomposition of $aA$ whose radical is $p$, say $q$, so $b\in q$ and thus it follows $b\in aA$.
To finish the proof you actually need to show that every Krull domain is completely integrally closed. This follows easily if you can prove that every DVR is completely integrally closed. Say $A$ is a DVR, and let's prove that $A$ is completely integrally closed. Consider $x\in K-A$ such that there is $0\ne a\in A$ with $ax^n\in A$ for all $n\ge1$. Then $x^{-1}\in A$. Let $t$ be a uniformizing parameter for $A$. Then $x^{-1}=ut^m$, where $u$ is invertible and $m\ge 1$. Then $x=u^{-1}t^{-m}$. Write $a=vt^r$, $v$ invertible and $r\ge0$. We get $ax^n=u^{-n}vt^{r-mn}\in A$ for all $n\ge1$, and this entails $t^{r-mn}\in A$ for all $n\ge1$, a contradiction.
