[Math] Is a compact set an union of a finite number of disjoint closed intervals

analysislebesgue-integrallebesgue-measurereal-analysis

I think it is true for $\mathbb R$ with usual metric. How about others? How to prove it?

Motivation:
I got this idea when I was reading a proof for Lebesgue's Criterion of Riemann Integrability, here it is:
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Please note the red marker: "[$[a,b] – \mathop {\cup}^{n}_{j = 1} I_j$] is a finite union of closed intervals". I think [$[a,b] – \mathop {\cup}^{n}_{j = 1} I_j$] is compact here. I've no idea how to prove the equivalence.

P.S. here is the definition of Lebesgue outer measure of Carothers' Real Analysis:
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Best Answer

No. See the Cantor set, which contains no proper interval.

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