Heine Borel: Are all closed and bounded sets generated by closed and bounded intervals? In these notes it is rumored to contain the most clear and intutive proof of Heine Borel  on the internet completely unmatched http://math.umaine.edu/~farlow/sec54.pdf 
I looked through it and found that they only proved it for the $[a,b]$ case (i.e. intervals) and left out the actual proof of the Heine Borel...
I wonder if proving $[a,b]$ directly implies that every closed and bounded set is compact. In other words,  are all closed and bounded set is generated by closed bounded intervals? 
 A: A closed bounded set is a closed set inside a big enough $[a,b]$. Since $[a,b]$ is compact and closed sets inside compact sets are compact, you have your result.
A: It depends on what you mean by generated. If you mean is every closed and bounded set the union of closed and bounded intervals, the answer is yes, because $\{x\}=[x,x]$ is a closed and bounded interval. But this is rather unsatisfying because then every set is generated by closed and bounded intervals. If you restrict yourself to finite unions of closed intervals, the answer is no (consider the Cantor set).
But once you have proves the Heine Borel Theorem for closed intervals, the rest is easy: Let $A$ be a closed and bounded set. Since it is bounded, it is a subset of $[a,b]$ for some $a$ and $b$. Let $\{G_a\}$ be an open covering of $A$. Then $\{G_a\}\cup \{[a,b]-A\}$ is an open cover of $[a,b]$. By the Heine Borel Theorem for intervals, there exists a finite subcover $\{G_{a_1},...,G_{a_n},[a,b]-A\}$ of $[a,b]$. Thus, $\{G_{a_1},...,G_{a_n}\}$ is a finite subcover of $A$.
Now suppose that $A\subset \mathbb R$ is compact. Cover it in sets of the form $(x-1,x+1)$ for each $x\in A$. Each of these intervals are bounded. Since $A$ is compact, this covering has a finite subcover, so $A$ is a subset of a finite number of bounded sets. Therefore $A$ is bounded. Now choose a point $y\not\in A$. For each $x\in A$, let $a_x=|x-y|/2$. The sets $(x-a_x,a+a_x)$ cover $A$, so there exists a finite subcover of sets $\{(x_1-a_{x_1},x_1+a_{x_1}),...,(x_n-a_{x_n},x_n+a_{x_n}).$  The intersection $$\bigcap_{i=1}^n (y-a_{x_i},y+a_{x_i})$$ is an open set that contains no points in $A$. Since $y\in A^C$ was arbitrary, we conclude that $A$ is closed.
