Confusion in proof of theorem ($2.7$) in Rudin's Real and complex analysis I am not able to fill the gap in proof of following theorem which is stated as...
Let $U$ be an open set in a locally compact hausdorff space $X$, $K\subset U$ and K is compact. Then there exists an open set $V$ with compact closure such that : 
$$K\subset V\subset \overline{V}\subset U$$
Proof : Let $p\in K$ as $X$ is locally compact there exists an open set $V_p$ such that $\overline{V_p}$ is closure. See that this collection $\{\overline{V_p}\}_{p\in K}$ is an open cover for $K$. as $K$ is compact this cover has a finite collection which covers $K$.
Suppose $V_1,V_2,\cdots,V_n$ covers $K$. I now set $V=\bigcup_{i=1}^n V_i$ see that $V$ is open being finite union of open sets and $K\subset V$
Now $\overline{A\cup B}=\overline{A}\cup \overline{B}$ so, for similar reasons we have  $\overline{V}=\overline{\bigcup_{i=1}^n V_i}=\bigcup_{i=1}^n \overline{V_i}$ and see that $\overline{V}$ is compact being finite union of compact sets.
So, I have an open set $V$ of $X$ with compact closure such that $K\subset V\subset \overline{V}$
Suppose my $U$ is $X$ then i am done as i need not check if $\overline{V}$ is in $X$
Suppose not then i am not sure if this closure is in $X$
Just to not get confused i now denote the union $\bigcup_{i=1}^n V_i$ as $G$
As $U^c\neq \emptyset$ I denote this $U^c$ as $C$.
I understood that given $p\in C=U^c$ there exists an open set $W_p$ such that $K\subset W_p$ and $p\notin \overline{W_p}$
I do not understand why $\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is empty..
Help me to fill this gap..
 A: I guess i got the solution and i thought it is not a better idea to edit the question..
So, I am writing this..
Suppose  $\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is non empty...
Then i have $x\in \bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$
In particular, $x\in C\cap \overline{G}\cap \overline{W_x}$
But then We have the condition that if $x\in C$ then $x\notin \overline{W_p}$
Thus we have a contradiction... so,$\bigcap_{p\in C}(C\cap \overline{G}\cap \overline{W_p})$ is empty.
A: For each $p \in C$ you have that $p \notin C \cap \bar{W_p}$  so in each term you have $p$ missing and if you take the intersection for $p \in  C$ then:
$$\cap_{p \in C}(C \cap \bar{G} \cap \bar{W_p}) \subset C \qquad p \notin (C \cap \bar{W_p}) \implies \cap_{p \in C}(C \cap \bar{G} \cap \bar{W_p}) = \emptyset$$
Also remember that $\bar{G}=\bar{V}$ so you can get the result that you're looking for. I don't know if I explained myself sufficently clear, if I haven't please let me know and I'll try to fix it.
A: By local compactness, for each $p \in K$, there exists an open neighborhood $V_p$ of $p$ with $\bar{V_p}$ compact and $\bar{V_p} \subset U$.
By compactness of $K$, there is a finite subset $F$ of $K$ for which $K\subset \bigcup_{p\in F} V_p$. Let $V = \bigcup_{p\in F} V_p$.
Certainly $V$ is open.
Finally, since $F$ is finite, $\bar{V} = \bigcup_{p \in F} \bar{V_p} \subset U$, and $\bar{V}$ is compact.
