Proving compactness of $\{0\}\cup\{\frac1n; n\in\mathbb N\}$ by definition Let $K\subset \mathbb{R}^1$ consists of $0$ and the numbers $1/n,$ for $n\in \mathbb{N}$. Prove that $K$ is compact directly from the definition (without using the Heine-Borel theorem)
Proof: Let $K$ is not compact. Then exists open cover $\{G_{\alpha}\}$ of $K$ such that contains no finite subcover. Then $K_1=K\backslash\{1\}$ cannot be covered by any finite subcollectio of $G_{\alpha}$. Then $K_2=K\backslash \cup_{i=1}^{2}\{\frac{1}{i}\}$ also. And we can proceed this process. It is clear that $\cap_{i=1}^{\infty}K_i=\{0\}$. It is clear that $0\in G_{\alpha}$ for some $\alpha$. But $G_{\alpha}$ is open then $\exists r>0:$ $|y|<r$ implies that $y\in G_{\alpha}$. We will take $n$ so large such that $1/(n+1)<r$. Then $K_n\subset G_{\alpha}$. If $z\in K_n=K\backslash \cup_{i=1}^{n}\{\frac{1}{i}\}$ then $|z|<\frac{1}{n+1}<r$ so $z\in G_{\alpha}$. We got contradiction because $K_n$ cannot be cover by any finite subcollection of $G_{\alpha}$
Is my proof true?  

I think that your proof is correct.
However why to proceed by contradiction? Consider an open cover. One of the open set $\mathcal O$ contains $0$. As $0 \in \mathcal O$, there is an open disk centered on $0$ included in $\mathcal O$. This disk contains all but a finite number of the points $1/n$. Consider the finite open sets of the cover that cover the remaining points plus $\mathcal O$. You have a finite cover of $K$.
