On the Spivak's proof of the theorem 3-11 (calculus on manifolds) In second paragraph of the case 1 within the proof:


*

*What is $U$ s.t $A\subset U$ and satisfies in the proof of the case 1 of theorem 3-11. 

*$\psi_i$ is defined on $U_i$ and its support is not compact. Now, How any $\psi_i$ is defined on $U$ s.t $\psi_1+\cdots+\psi_n$  to be well-defind on $U$.

*Why is $\psi_1+\cdots+\psi_n>0$ on $U$?


 A: For your third question: the bottom of page 63 says that the $\psi_i$ are nonnegative functions, so the only question is why at least one of them is positive at each point. Well, since the interiors of $D$ cover the space, every point is in some $D_i$, and $\psi_i$ is positive on $D_i$, so you're done. 
For question 2, since the (finite) sum of the $\psi$s is positive, you can divide by it; the only problem would be if it were zero someplace. So $\phi_i$ is well-defined. If your concern is "what's the value of $\psi_i$ outside of $U_i$?", the answer is that it's defined to be zero out there, so the sum of the $\psi$s is actually defined everywhere. 
I don't understand question 1. 
A: Here are some of my thoughts for clarification: 
(1) Yes, we can take $U:= \bigcup \mathring{D_i}$. But more generally we can take the interior of the set of points such that $\sum \psi_i =0$. This contains $\bigcup \mathring{D_i}$ by construction.
(2) As $A \subseteq U$, $\phi_i := \frac{ \psi_i}{\sum \psi_i } : U  \rightarrow \mathbb{R}$ is well defined, since $x \in D_i$ for some $i$ making $\psi_i > 0$. Note also that $\phi_i \in C^{\infty}(U)$ by Chain Rule. 
(3)  It suffices to show $supp (\phi_i) \subseteq U_i$ for each $i$. I believe the text is phrased weirdly (someone please correct me if I am wrong) in the start:

...$\phi$ is defined on an open set containing $A$...

This seems to imply $\phi$ has domain $U$ open, so subordinate condition is satisfied as $supp(\phi_i) \subseteq supp(\psi_i)$ and we are done. I believe we need $\{ \phi_i : \mathbb{R}^n \rightarrow \mathbb{R}\} $ in $C^{\infty}(\mathbb{R}^n)$, as given in this general definition, with $X = \mathbb{R}^n$. 
(4) So our last step is to extend domain. The problem is whether we can define $\phi_i(x) = 0$ for all $x \in [supp(\psi_i) \cap U]^c$. But this does not work! 
This is best seen if we only have one open set $U_1$ which covers $A$. We obtain $\phi_1 (x)=1$ on the open set $U = \mathring{D_1}$ that contains $A$. So if we define $\tilde{\phi_1(x)} = 0$ for all $x \notin supp(\psi_1) \cap U = U$ our function is no more continuous (by construction $\phi_1(x) > 0$ on $D_1$). 
(5) To resolve the problem in highlighted above, $A$ is compact subset of $U$ so we have $f \in C^{\infty}(\mathbb{R})$ such that $supp (f) \subseteq U$. By defining 
$$ \tilde{\phi_1}(x) = \begin{cases} f \cdot \phi_1(x) & x \in U \\ 
0 & x \notin U \\ 
\end{cases} 
$$
we obtain a $C^{\infty}(\mathbb{R}^n)$ function.  
A: I recommand that $U:=\cup_1^n \text{interior}\ D_i$.
Now for compact set  $D_i\subset U$, there exist non negative $C^{\infty} $ function  $\psi_i: U\rightarrow [0,1]$  which is positive on $D_i$ and 0 outside of some closed set ( i.e. $\text{supp}(\psi_i)$) contained in  $U$.
