$\quad$$\quad$Let a subset $S\subseteq \mathbb{R}^{3}$ is a regular surface.(that is to say,for each $\mathbb {x}\in S$,there exists an open set $U\subseteq \mathbb{R}^{2},$ and open neighborhood $V$ of $\mathbb {x}\in \mathbb{R}^{3}$,and a surjective continuous function $\varphi:U\rightarrow V\cap S$ such that $\textit {1.} \varphi$ is continuously differentiable; $\textit {2.} \varphi$ is a homeomorphism;$\textit {3.}$ For each $(u,v)\in U$, the differential $d\varphi_{(u,v)}:\mathbb{R}^{2}\rightarrow \mathbb{R}^{3}$ is a one-to-one linear transformation.)$\quad$
$\quad$$\quad$Define $\partial S=\overline{S}\backslash S$ (We can think of $\partial S$ as the boundary of the regular surface $S$). Let$K_{n}=\{\mathbf{x}\in S:|\mathbf{x}|\leq n,dist(\mathbf{x},\partial S)\geq \frac{1}{n}\},$ here $dist(\mathbf{x},\partial S)=\inf\{|\mathbf{x}-\mathbf{y}|:\mathbf{y}\in \partial S\}.$Then$S=\bigcup_{n=1}^{\infty}K_{n},$and that each $K_{n}$ is closed and bounded in $\mathbb{R}^{3}.$$\qquad$$\qquad$
$\quad$$\quad$Since $S$ is a regular surface, choose for every $\mathbf{x}\in S$ an open set $U_{\mathbf{x}}\subset \mathbb{R}^{2},$ an open subset$V_{\mathbf{x}}\subset S,$ and a $C^{1}$ map $\varphi_{\mathbf{x}}:U_{\mathbf{x}}\rightarrow V_{\mathbf{x}}$ that is injective with injective derivative.Choose $B_{\mathbf{x}}\subset U_{\mathbf{x}}$ an open ball with $\overline {B_{\mathbf{x}}}\subset U_{\mathbf{x}}$ and set $V^{'}_{\mathbf{x}}=\varphi_{\mathbf{x}}(B_{\mathbf{x}}).$$\quad$$\quad$
$\quad$$\quad$Show that we can choose a sequence $\mathbf{x}_{1},\mathbf{x}_{2},\cdots$ and integers $p_{1} \leq p_{2}\leq\cdots$ such that $\bigcup_{i=1}^{p_{n}}V^{'}_{\mathbf{x}_{i}}$ is an open cover of $K_{n}$,and that we can choose the $U_{\mathbf {x}_{i}}$ all disjoint, and such that only finitely many intersect any ball.
$\quad$$\quad$ I'm really puzzled that words $\textit{"and that we can choose the $U_{\mathbf {x}_{i}}$ all disjoint,}$ $\textit{and such that only finitely many intersect any ball."}$ Actually,it took me lots of time to prove it.Until now, there is no progress at all. What role does it play? How to deal with this problem? Your useful help will be greatly appreciated！
That role that it plays can only be judged if we know what is done with the $U_{\mathbf {x}_{i}}$. So let us see how to get them.
Although it is not really important, let us note that it is possible that $\partial S= \emptyset$. In that case $K_n = S \cap D(0,n)$ where $D(0,n)$ denotes the closed ball with center $0$ and radius $n$.
As you know, each $K_n$ is compact. We can therefore find finitely many $\varphi^n_{\mathbf {x}^n_{i}} : U_{\mathbf {x}^n_{i}} \to V_{\mathbf {x}^n_{i}}$, $i = 1,...,k(n)$, such that $K_n \subset \bigcup_{i=1}^{k(n)} V'_{\mathbf {x}^n_{i}}$. Now set $p_r = \Sigma_{n=1}^r k(n)$ and arrange the $\mathbf {x}^n_{i}$ to a sequence via $\mathbf {x}_{i} = \mathbf {x}^r_{i - p_{r-1}}$ for $p_{r-1} < i \le p_r$. Hence, for each $i$ there exist a unique $r(i)$ such that $\mathbf {x}_{i} = \mathbf {x}^{r(i)}_{i - p_{r(i)-1}}$
Define $U_i = U_{\mathbf {x}^{r(i)}_{i - p_{r(i)-1}}}$, $V_i = V_{\mathbf {x}^{r(i)}_{i - p_{r(i)-1}}}$ and $\varphi_i = \varphi_{\mathbf {x}^{r(i)}_{i - p_{r(i)-1}}}$. The $U_i$ are open subsets of $\mathbb{R}^2$, but they need not be pairwise disjoint. Let us adjust them. There is a diffeomorphism $h : B(0,1) \to \mathbb{R}^2$, where $B(a,r)$ = open ball with center $a$ and radius $r$ (take for example $h(x) = \frac{x}{1 - \lVert x \rVert^2}$). Moreover, the translations $t_i : \mathbb{R}^2 \to \mathbb{R}^2, t_i(\mathbf{x}) = \mathbf{x} + (2i,0)$ are diffeomorphisms.
Replace $U_i$ by $U'_i = t_i(h^{-1}(U_i))$ and $\varphi_i$ by $\varphi'_i = \varphi \circ g_i$, where $g_i = h \circ t_{-i} : U'_i \to U_i$.
Then the $U'_i$ are pairwise disjoint because they are contained in $B((2i,0),1)$. Moreover, each ball $B$ is contained in some $B(0,2i)$ and $U'_j \cap B \subset B((2j,0),1) \cap B(0,2i) = \emptyset$ for $j > i$.