The Borel–Lebesgue property

Definition. $K$ is compact if every open cover of $K$ admits a finite subcover.

Proof sketch for $[0,1]$. Suppose $(U_i)$ is an open cover with no finite subcover. Bisect $[0,1]$: one half also has no finite subcover. Iterate to get nested intervals $I_n$ of length $2^{-n}$, each requiring infinitely many $U_i$. Their intersection is a single point $x\in[0,1]$, which lies in some $U_{i_0}$. For $n$ large, $I_n\subseteq U_{i_0}$ — contradiction.

Conclusion. $[0,1]$ is compact.