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I had to give an example showing subspace of compact space is not compact. For this I first picked example of compact space from the book Topology by James R. Munkres section 26, solved example 2.

$X= \{0\}\cup\{\frac{1}{n} | n\in \mathbb N \}$. This space $X$ is compact space as subspace of $\mathbb R$. Now Let, $Y= \{\frac{1}{n}\mid n\in \mathbb N\}$ I claim that $Y$ is not compact as subspace of $X$. To show this I used the fact that if $Y$ is subspace of $X$ then $Y$ is compact iff every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$. First I showed that each singleton set $\{\frac{1}{k} \}$ is open in $X$. Then, $Y=\bigcup_{k=1}^{\infty} \{\frac{1}{k} \}$ So, this collection is open cover of $Y$ by sets open in $X$. Hence, by above fact if $Y$ is compact then there exist a finite subcollection of above singleton sets. But, this is not possible because, if such finite subcollection exist then $Y$ would be finite but, $Y$ is infact infinite set. So, we conclude that $Y$ is not compact set.

Is my solution is correct? If not, then please tell me corrections.

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  • $\begingroup$ Sure, that works. $\endgroup$
    – BrianO
    Jun 5 '20 at 8:33
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This is a correct solution.

In general, as long as you pick a closed and bounded subset of $\mathbb{R}^n$ (such as a closed ball), and take a non-closed subset (e.g open ball), it satisfies the requirements of the question. Your example is similar.

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  • $\begingroup$ Thank you so much $\endgroup$ Jun 5 '20 at 9:35

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