Show there is a countable collection of closed balls $\beta=\{B_n=\overline{B_{r_n}(x_n)}:n\in \mathbb{N}\}$ such that $U=\cup_nB_n^{\text{int}}$ 
Let $U= \cup C$ be an open set in $\mathbb{R}^n$. Show there is a countable number of closed balls $\beta=\{B_n=\overline{B_{r_n}(x_n)}:n\in \mathbb{N}\}$ such that:
a.$U=\cup_n B_n^{\text{int}}$
b. For each $n\in \mathbb{N}$ there exists an open set $M$ such that $B_n\subseteq M$.
c. For every point $x\in U$, there exists a neighbourhood $V$ of $x$ such that only finitely many elements of the cover $C$ intersect $V$ non trivially.

So I want to use compact exhaustion of the open set $U$.
Fix a compact exhaustion $K_n:n\in\mathbb{N}$ of $U$
Then since by definition for every $i\in\mathbb{N}$ $K_i\subseteq K_{i+1}^{\text{int}}$
I can make a compact set $L_n=K_n\setminus K_{n-1}^{\text{int}}$ which will satisfy that each $L_i$ is compact as its an intersection of closed sets, which will be closed and is bounded by $K_i$.
From here I think what I want to do is to take balls $B_i(x)$ for each $x\in L_i$ and take a radius which will not intersect more then $L_{i+1}$ or $L_{i-1}$. A set of all such balls $B_n$ in each $L_i$ will admit a finite subcover because each $L_i$ is compact. And the union of all such finite subcovers of each $L_i$ should still cover $U$ and because $L_n$ is countable this set $\cup B_n$ will be countable.
I believe this would satisfy part a. but I don't really see part b and c. I also don't know what a non trivial intersection is.
 A: Take a family $V=\{V_n:n\in \Bbb N\}$ of bounded open sets such that $U=\cup V$ and such that $\overline V_n \subset V_{n+1}$ for each $n.$ (To get $V,$ see APPENDIX below.)
Let $K_1=\overline V_1.$ For $n>1$ let $K_n=\bar V_n \ V_{n-1}.$ Each $K_n$ is compact.
Let $C_1$ be a finite cover of $K_1$ by open balls, such that $\overline {\cup C_1}\subset V_2.$ This is possible because $K_1$ is compact, $V_2$ is open, and $K_1\subset V_2.$ 
For $n>1$ let $C_n$ be a finite cover of $K_n$ by open balls, such that $\overline {\cup C_n}\subset V_{n+1}$ and also such that $\cup C_n$ is disjoint from $\cup_{j\le n-2}(\overline {\cup C_j}).$
The last condition above is vacuous when $n=2.$ When $n>2$ it is crucial,  and  possible because  $S_n=\cup_{j\le n-2}(\overline {\cup C_j})$ is a subset of $\cup_{j\le n-1}V_j=V_{n-1},$ and $V_{n-1}$ is disjoint from $K_n,$ so the closed set $S_n$ and the compact set $K_n$ are disjoint. 
Now $C=\cup_{n\in \Bbb n}C_n$  is a countable family of open balls, and $U=\cup C=\cup _{b\in C}\, int(\bar b)=\cup_{b\in C}\,\bar b.$
If $p\in U$ then for some $n$ we have $p\in V_n\subset \overline V_n=\cup_{j\le n}K_j$ so for some $j\le n$ and some $b \in C_j$ we will have $p\in b\in  C_j.$ Now this $b$ is disjoint from $\cup C_m$ for every $m\ge n+2.$ So $\{b'\in C: b'\cap b\ne \phi\}$ is a subset of the finite set $\cup_{j\le n+1}C_j.$
APPENDIX. Let $\bar 0$ be the origin in $\Bbb R^n.$
(i). If $U=\Bbb R^n$ let $V_n=B_n(\bar 0)$ for each $n\in \Bbb N.$
(ii). If $U\ne \Bbb R^n,$ let $U^c=\Bbb R^n$ \ $U$ and let $d$ be a metric for the topology on $\Bbb R^n,$ and for $p\in U$ let $d(p,U^c)=\inf \{d(p,q):q\in U^c\}.$
Now let $V_1=\{p\in U\cap B_1(\bar 0): d(p,U^c)>1\}.$ And for $n>1$ let $V_n=  \{p\in U\cap B_n(\bar 0): d(p,U^c)>1/n\}.$ I will leave it to the reader to confirm that $\{V_n:n\in \Bbb N\}$ has the required properties. 
A: Here's an answer to part a:  This is just the same as, or a slight modification of the proof that was posted earlier. 
For any set $U$ the set of balls $$\mathcal{B}=\{B_p(q)^{\mathrm{int}}~|~q\in \mathbb{Q}^n, p \in \mathbb{Q}_{\ge 0}\}$$ form a countable cover of $U$. 
Let $x$ be an arbitrary point in $U$. Then $x$ is contained in an open ball $B_{r_x}(x) \subseteq U$. Let $q$ be a rational point contained within the ball $B_{r_{x/3}}(x)$. Then we can choose a ball $B_p(q) \in \mathcal{B}$ with $r_x/3 < p <2r_x/3$. Then $$x \in B_p(q) \subset B_{r_x}(x) \subseteq U.$$ 
Hence $U$ is the union of a subset of $\mathcal{B}$.
