Proving Heine-Borel using Completeness Axiom I think the way to approach this is to take U, a subset of R.  Let U be bounded.  Then by the Completeness Axiom, there exists a supremum and an infinum of U.  I think its obvious that these are part of U, but I don't know how to show it.  This would give that U is closed.  Then take an open cover of U.  I have no idea how it has a finite subcover
 A: Let's first handle the case of a closed interval $[a, b]$. Let $\mathcal{C} $ be an open cover of $[a, b] $. And therefore there is an open interval $I\in\mathcal{C} $ such that $a\in I$. Consider the set $$A=\{x\mid x\in[a, b], [a, x]\text{ is covered by a finite number of intervals from }\mathcal{C} \} $$ Then from the above argument $A$ is non-empty and obviously bounded above. Hence by Completeness Axiom $c=\sup A$ exists and $a<c\leq b$. We will prove that $c=b$ and $c\in A$.
Note that there is an interval $J\in\mathcal{C} $ such that $c\in J$. And since $c=\sup A$, it follows that there is an element $x\in A$ such that $x\in J$. Now $[a, x] $ is covered by a finite number of intervals from $\mathcal{C} $ and including the interval $J$ we now see that $[a, c] $ is covered by a finite number of elements from $\mathcal{C} $. Thus $c\in A$.
Next let us assume on the contrary that $c<b$. As before we have an interval $J\in\mathcal{C} $ such that $c\in J$ and clearly we have an $x\in J$ such that $c<x<b$. Moreover $[a, c] $ is covered by a finite number of intervals from $\mathcal{C} $. Adding $J$ to this finite collection of intervals we see that $[a, x] $ is also covered by a finite number of intervals from $\mathcal{C} $. Thus $x\in A$. But this is an obvious contradiction as $x>c=\sup A$. It follows that our assumption is wrong and $c=b$. Thus $[a, b] $ is covered by a finite number of intervals from $\mathcal{C} $. 
You should now try to prove the same for a general closed and bounded subset $S$. I will provide some hints here. If $S$ is finite then we are done. Let $S\subseteq[a, b] $ and $\mathcal{C} $ be an open cover for $S$. and consider any point $x\in[a, b] \setminus  S$. Then $x$ is not an accumulation point of $S$. Hence there is an open interval $I_{x}$ such that $x\in I_{x} $ and $I_{x} \cap S=\emptyset$. Add all such intervals $I_{x} $ corresponding to $x\in[a, b] \setminus S$ to the cover $\mathcal{C} $ to obtain an open cover $\mathcal{D} $ of $[a, b] $. Now use the case of Heine Borel Theorem already proved. 
