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Counterexample: $[0,1]\times[0,1]$ with induced subspace topology from $\mathbb{R}^2$ is compact. The open cover $\mathscr{U}$ is just the two circular sectors. When we look at the up-left corner and down-right corner, it fails - there is no such $\delta$, such that let the open ball be only in one of circular sectors... what is wrong?

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  • $\begingroup$ There is no way to make those two quarter disks open and to include the corners $(1,0)$ and $(0,1)$. $\endgroup$ – Thomas Andrews Oct 9 '17 at 23:27
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The upper-left and bottom-right corners are not contained in either open circle. That is, your claimed counterexample fails to cover the square.

Moreover, I think you may have misunderstood what is important about Lebesgue's number lemma. The following statement

Let $\mathcal U$ be a cover. For any $x$, there is some $\delta$ and some $U\in \mathcal U$ such that $B(x,\delta)\subseteq U$

is clearly true of any cover - no need for compactness. This is basically just writing down what "open" and "cover" are defined to be. Given that your counterexample fails this condition, it is clear that it is not an open cover.

Note that Lebesgue's number lemma exchanges the quantifiers and says that there is some $\delta$ that works for all $x$. This is what requires compactness.

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  • $\begingroup$ Thanks for pointing out. I did not thought in that way, I though the lemma was weirdly wrong, because when I fix a $\delta$ and then moving a small ball towards that two corner, there must be a time point that the small ball is not fully contained in any circular sectors. But yes.. there are another two sectors missing to cover the two corners. $\endgroup$ – Upc Oct 9 '17 at 23:39

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