Every metric space is first countable, and that makes countably compact and sequentially compact equivalent. We only need show compact = countably compact.
Assume S is countably compact. Equivalently, every sequence has a cluster point. We're going to prove a lemma about finite sets, then we will show S is compact.
Suppose, for a given ε, every finite set of points remains farther than ε from some point in S. As long as the set is finite, we can't get close to some parts of S. Build a sequence p1 p2 p3 etc, such that each point is at least ε away from all the earlier points. If x is a cluster point of this sequence, than there are infinitely many points within ½ε of x, which is impossible. We have built a sequence with no cluster point, and that contradicts the fact that S is countably compact. Therefore every ε has a finite set of points, such that all of S is within ε of the points in that finite set.
Set ε = 1/n, and find a finite set Wn for ε. Let U be the union of these finite sets Wn. Note that U is a countable set. Furthermore, the center of an open ball is always close to something in U. In other words, every open ball intersects U, and U is a dense set. Thus S is separable, hence second countable. In a second countable space, compact and countably compact are equivalent.
That completes the proof. In a metric space, compact = countably compact = sequentially compact.
Next, cover S with finitely many balls of radius ½. This means there are finitely many intersections between balls of radius 1 and balls of radius ½, and these intersections cover all of S. In particular, the intersections of the balls of radius ½ with B1 cover B1. Choose B2 so that B1∩B2 contains infinitely many points of p.
Cover S with finitely many balls of radius ¼, and choose B3 such that B1∩B2∩B3 contains infinitely many points of p. continue this process, as the radius approaches 0.
Build a subsequence of p as follows. Start with any point of p that is contained in B1. Then select a later point of p in both B1 and B2. Then a third point of p from B1∩B2∩B3, and so on. Show that this subsequence is cauchy. Since S is complete, let the subsequence converge to x. This proves S is sequentially compact, hence compact.
Select an irrational point q in the interval. Let On be the open set [0,b)∪(c,1], where b and c are carefully chosen rational numbers bracketing q, such that c-b < 1/n. This is an open cover with no finite subcover.
If S is closed in Rn it is complete. If S is bounded it is totally bounded. This makes S compact.
Conversely, let S be compact in Rn. If S is not bounded, ever expanding balls cover S, and there is no finite subcover. This is a contradiction, thus S is bounded.
If S does not contain one of its limit points x, build a sequence that approaches x, and S fails to be complete. Since compact implies complete, S has to be closed.
In Rn, compact is equivalent to closed and bounded. A line segment, a square, a circle, and an n dimensional hypercube; these are all compact sets.