Show that the compact set $X$ is contained in union of closed sets contained in open cover. Let $X$ be a compact set of $\mathbb{R}^{n}$, and let $\mathcal{O}$ be an open cover of $X$.
a) Show that $X \subset B_{1} \cup \cdots \cup B_{k}$, for some $k$, where each $B_{i}$ is a closed ball lying in some element $U = U_{i}$ of $\mathcal{O}$.
b) Prove there exists $\epsilon > 0$ such that , for all $x \in X$, $B_{\epsilon}(x) \subset U$ for some $U \in \mathcal{O}$.
The solutions I'm presenting are not the full written out formal solutions, but the "idea" and reasoning behind the solution. 
Attempted Solutions:
a) Trying to reword the question in a form to make it more understandable for myself: We want to show: $X$ is contained in a finite union of closed sets.
$X$ is compact. This implies there exists a finite subcover for every open cover $\mathcal{O}$ of $X$. It also means that each element $x_{i} \in X$ is also an element of an open set $U_{i}$, i.e $x_{i} \in U_{i}$.
By Heine Borel we also know that $X$ is closed and bonded so each $x_{i}$ is in the closesed set $X$.
Consider the intersection $X \cap U_{i}$. Each of these sets, consist only of the points $x_{i} \in X$. That is $X \cap U_{i} = \{x_{j}\}_{i} , 1 \leq j \leq n$. Where $\{x_{j}\}_{i}$ is just the set of points corresponding to that respective intersection of $U_{i} \cap X$.
A finite collection  of points is a closed set. Therefore there exists a closed set $B_{i}= \{x_{j}\}_{i} \subset U_{i}$. By taking the union of these $B_{i}$ we have our set $$, that is $X = \bigcup_{i = 1}^{n}B_{i}$.
Comments: I was trying to think of possible ways of how I can break my proposed solution, but nothing came to me yet. Perhaps I'm missing something in how I'm thinking about thing. I'm on the fence about it. I feel like the solution is "right", but I also have a feeling in my gut that it isn't.
b) We want to show that every point $x \in X$ has a ball of radius $\epsilon >0$ such that $B_{\epsilon}(x) \subset U_{i}$.
Given that $X$ is compact, then there exists a finite subcover. That means for all $x_{i} \in X$, that $x_{i} \in U_{i}$ where $U_{i}$ is a part of the finite open cover such that $\bigcup_{i = 1}^{n} = X$. 
The definition of a set $Y$ being open is that for every point $y_{i} \in Y$ there exists a ball of radius $\epsilon >0$ such that $B_{\epsilon}(y_{i}) \subset Y$.
Therefore by definition there exists $\epsilon >0$ such that $B_{\epsilon}(x_{i}) \subset U_{i}$
Comments: Should I be providing an explicit $\epsilon$? and if so how? Just because these questions were joined together I feel that I should be using something from part a) to prove this fact. 
Guidance and suggestions for both of these questions ?
 A: a) Instead of taking a single point, try taking a ball whose closure lies in an element of $\mathcal O$, then get a finite cover by compactness. b) You don't have to find an explicit $\epsilon$. You just have to show that such an $\epsilon$ exists.
As for using (a) to prove (b), it doesn't have to be the case. (b) is actually a well-known result and the proofs I'm familiar with don't use (a).
The following is a more concrete hint for (a): For each $x \in X$, take $B_r(x) \subset U$ for some $r > 0$ and $U \in \mathcal O$. This is possible because $\mathcal O$ is an open cover of $X$. Now for each ball, take a smaller ball that lies in the original ball. Show that it's still a cover. Get a finite subcover, take closure, and show that the result satisfies the desired property. 
A: A.  Since X is regular, for all x in X,
exists U in O, open V, closed K with x in V, V subset K, K subset U.
As the V's constitute an open cover, finite many will cover X.
Thus finite many of those K's having the desired property will cover X.    
