I think the answer is no, here is my attempt at a proof:

Consider the sequence of points in $[0,1]$, $\{x_n\}_{n=1}^\infty$, with $x_n=n^{-1}$. Let $F_1 = \{x_n\}_{n=1}^\infty$, $F_2=\{x_n\}_{n=2}^\infty$, etc, with $F_k=\{x_n\}_{n=k}^\infty$ for $k\geq 1$. Then, since each $F_k$ is countable for $k\geq 1$, the complements $U_k=\mathbb{R}\backslash F_k$ are open. The union, $\bigcup\limits_{n=1}^\infty U_n$, contains $[0,1]$ and thus is an open cover for $[0,1]$. Consider a finite union of the sets $U_k$, e.g. $\bigcup\limits_{k=1}^K U_{n_k}$ where $K$ is a positive integer. Then, let $N=\max\limits_{1\leq k \leq K}n_k$. The point $(N+1)^{-1}$ is not in the finite union but is in the set $[0,1]$ so any finite union is not a cover. Thus $[0,1]$ is not compact as there is at least one open cover which does not admit a finite open subcover.

  • 3
    $\begingroup$ Yeah, that's just it $\endgroup$ – Gabriel Sanfins Feb 22 '17 at 4:59

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