Covers, subcovers, and the definition
An open cover of a space X is a collection of open sets whose union is all of X — every point is caught by at least one set. A finite subcover is a finite handful chosen from that collection that still covers X. The space X is compact if every open cover admits a finite subcover. Read it as a challenge–response game: an adversary hands you any open cover, however infinite or clever, and you must always reply with finitely many of its members that already do the job.
Heine–Borel and a proof for [0,1]
The Heine–Borel theorem is the bridge to ordinary intuition: a subset of Rⁿ is compact if and only if it is closed and bounded. Boundedness alone fails — (0, 1) is bounded but not compact — and closedness alone fails — all of R is closed but not compact. You need both. The heart of the theorem is that any closed bounded interval is compact; the proof uses the completeness of R in an essential way.
Theorem. [0, 1] is compact.
Let C be an open cover of [0, 1]. Define
S = { x in [0,1] : [0, x] can be covered by FINITELY many sets of C }.
We show 1 ∈ S, which says [0,1] = [0,1] has a finite subcover.
Step 1 (S nonempty). 0 ∈ [0,1] lies in some open U0 ∈ C, so {U0} covers
[0,0] = {0}. Hence 0 ∈ S, and S is nonempty and bounded above by 1.
Step 2 (let s = sup S). By the least-upper-bound property s exists,
and 0 ≤ s ≤ 1. Some open U ∈ C contains s; openness gives a small
interval (s - d, s + d) ⊆ U for some d > 0.
Step 3 (s ∈ S). Pick a point t with s - d < t ≤ s and t ∈ S
(possible since s = sup S). Then [0, t] has a finite subcover F.
Adding U covers [0, t] ∪ (s - d, s + d) ⊇ [0, s].
So F ∪ {U} is a finite cover of [0, s], giving s ∈ S.
Step 4 (s = 1). If s < 1, the SAME finite set F ∪ {U} already covers
[0, s'] for some s' with s < s' < min(s + d, 1), so s' ∈ S — contradicting
that s is an upper bound of S. Hence s = 1, and 1 ∈ S by Step 3.
Therefore [0,1] has a finite subcover of C. Since C was arbitrary,
[0,1] is compact. ∎Sequential compactness
There is a second, more hands-on notion. A space is sequentially compact if every sequence in it has a subsequence converging to a point of the space. In Rⁿ — and in every metric space — this is equivalent to compactness. For [0, 1] the equivalence is just the Bolzano–Weierstrass theorem: every bounded sequence has a convergent subsequence, and closedness keeps the limit inside. The two definitions feel different — covers versus sequences — yet describe exactly the same metric spaces.