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I am trying to self study real analysis , and that is when I cam across this doubt .

The way I understand Heine Borel theorem is that , whenever A is contained in union of open sets , then A is contained in the union of finitely many of these sets.

Formally ,if $A \subset\bigcup\limits_{i\in I} G_{i}$ and each $G_i$ is open ,then there exist $i_1,i_2,...i_N$ such that $A\subset\bigcup\limits_{i=1}^{i=N} G_{i}$

Now my doubt arose when I was asked to show that B(x,r) is not compact. ( I get that I could use the later result of using that a compact set has to be closed and bounded to prove it, but I just wanted to work with the Heine Borel theorem here) One of the main reason that I do not get why this set has to be compact is the fact that I don't understand why can't I make a open ball B(x,R) where R>r , which will be an open cover for B(x,r) and obviously then we would have found out a finite open cover for B(x,r) thus making it a compact set ! ( It is safe to imagine that we are talking of the case where the ball exists in $R^n$ metric space )

I know that this doubt should be trivial but for some stupid reason I am just not able to understand this.

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    A city is called friendly if every single person living in it loves pasta. San Diablo has a house for sale. I tell Freddy, who loves pasta to buy it and move in. Now I claim that San Diablo is friendly because it has a person in it who loves pasta. Do you see why I am wrong. You did the exact same thing in showing that *one* open cover has a finite subcover. You need to show *all* open covers have a finite subcover. Moving a friendly one in won't do it. – 2017-02-16

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In order for a set to be compact, any open cover must have a finite subcover. You have only demonstrated that one open cover has a finite subcover (in fact the cover itself is finite).

For instance, let $U=B(0,r)$ the open ball, and define an open cover of $B(0,r)$ $$\bigcup_{n=1}^\infty B(0,r-1/n).$$ You can show this doesn't have a finite subcover, cause if $n=N$ is the index of the largest ball used in the finite subcover then the union of the subcover is equal to $B(0,r-1/N)$ and thus does not cover $B(0,r).$

We have demonstrated there is an open cover with no finite subcover and thus the open ball is not compact.

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    Would you kindly explain how $B(0,r)\subseteq\bigcup\limits_{n=1}^\infty B(0,r-\frac1n)$ if $B(0,r)\nsubseteq B(0,r-\frac{1}{n})$, $n\rightarrow \infty$? – 2017-10-14
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    @Diracology Let $x\in B(0,r).$ The distance between $0$ and $x$ is less than $r$ so write it $d(0,x) = r-\epsilon$ for some $\epsilon>0. $ Then $x\in B(0,r-1/n)$ for all $n>1/\epsilon$ so $x\in \bigcup_{n=1}^\infty B(0,r-1/n).$ – 2017-10-14
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    But if $x\in B(0,r-1/n)$, $n>1/\epsilon$, and we have chosen $x\in B(0,r)$ with the largest distance $d(0,x)$ then why $B(0,r)\nsubseteq B(0,r-\frac{1}{n})$, $n>1/\epsilon$? I have never had formal studies in analysis or topology so I am trying to visualize this example and I am missing something. But I do not see that. – 2017-10-14
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    @Diracology I'd just imagine (or draw) a sequence of open intervals approaching $(-r,r).$ For any fixed $n$ we can always find a point in $B(0,r)$ that is outside $B(0,r-1/n),$ Just take one with distance greater than $r-1/n$ (and less than $r$... we know there are numbers in between there). So $B(0,r)\not\subset B(0,r-1/n).$ But for every $x\in B(0,r)$ there is an $n$ such that $x\in B(0,r-1/n).$ When we fix $n$ it's never enough to cover the whole of $B(0,r)$ and yet then union of *all* the $B(0,r-1.n)$ does cover every point. – 2017-10-14
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    Thanks, I think I got it now. Perhaps my mistake was to think of a point inside the open disc with the greatest distance to the origin. But this would not make sense, there is always a point inside the same open disc with greater distance. – 2017-10-14
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    @Diracology Yep, that's right. Openness is key here. – 2017-10-15
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For $n \in \mathbb N$ with $n>r$ let $B_n=B(x, r-\frac{1}{n})$. Then the sets $B_n$ are an open cover of $B(x,r)$, which contains no finite subcover of $B(x,r)$