Showing $S^n$ is compact The definition of compact is every open cover has a finite subcover; I want to prove $S^n$ is compact directly, i.e. choose any infinite open cover, $\{a_i\}$, how to find a finite subcover?  
What spaces can we get if we only require that at least one infinite open cover has a finite subcover,by proper open subsets., is this sufficient for compactness?
Also, how to show (closed and bounded) $\implies$ compact?
How to prove that every open cover of [0,1] has a finite subcover?
 A: For the first question, proving $S^n$ is compact directly from the definition could be very long and messy; I would suggest you use Heine Borel (your third question): consider the Euclidean norm $$ | \ \ \  |: \mathbb{R}^n \to [0, +\infty)$$ this is a continuous function, hence the preimage of a closed set of $[0, +\infty)$ is closed in $\mathbb{R}^n$. Since points are closed in $\mathbb{R}^n$ $$S^n = | \ \ \ |^{-1} (1) $$
is closed. Since the $S^n$ is bounded, it is compact. 
For the second question the answer is all spaces with infinite open sets. Take a general topological space $(X, \tau)$, and simply take the cover of all open sets, $\tau$: a finite subcover is trivially given by $\{ X \}$. Also, asking that $X$ not be part of the cover still leaves you quite far from compactness: take $(0, 1)$, and consider an open cover $\mathscr{U}$ such that $\left(0, \frac23 \right), \ \left(\frac13, 1 \right) \in \mathscr{U}$. (If you want $\mathscr{U}$ to be infinite, just throw in infinite other open sets). $\{\left(0, \frac23 \right), \ \left(\frac13, 1 \right) \} \subseteq \mathscr{U}$ is a finite subcover of $\mathscr{U}$, and clearly an open interval is far from being compact.
For the third question I suggest you read here , or here. Also note that the " $\Leftarrow $" is also true.
For your last question, I will include a sketch of a proof I am familiar with (the whole proof is rather long, and you can find it, and probably others, in many General Topology textbooks): 
Let $I = [0, 1]$, let $\mathscr{U}$ be an open cover of $I$, and let $$X = \{x \in I \ | \ [0, x] \ \text{is covered by finitely many}  \ U \ \in \mathscr{U}  \}$$
if you prove that $I = X$ you are done. You can do this by proving that $ \emptyset \neq X$ is an interval, and is simulteneously open and closed in $I$. Since the only interval $\subseteq I$ with this property is $I$ itself, you are done.  
(sorry, I'm slow with latex and by the time I finished, most of this was already covered in the comments; I hope it helps anyway!)
