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I'm asked to prove that $\{0,1, 1/2, 1/3, \dots \}$ is a compact set without the help of Heine-Borel Theorem. I have to use only the definition of a compact set.

$\textbf{My work}$:

As long as the definition of a compact set says that $K$ is compact if for every open covering of $K$ there is a finite subcovering (is this the right word in English?) of $K$, then I started taking an arbitrary open covering, say $\mathcal{G} = \{G_{\lambda}: \lambda \in L\}$. Well, now I have to prove that there is a finite subcovering. If $A = \{0,1, 1/2, 1/3, \dots \} \subseteq \bigcup G_{\lambda}$, then every element $a$ of $A$ is such that $a \in G_{\lambda}$ for some $\lambda$. But I'm stuck at this point. Any ideas/hints? Thanks!

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    Every neighborhood of $0$ contains all but finitely many points of your set (prove this). – 2017-01-25
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    Technically, with your setup $A\subseteq \bigcup \mathcal{G}$ – 2017-01-25
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    sure @user160738! Just edited, thanks! – 2017-01-25

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Hint: at least one open set of the cover contains $0$. How many elements of the set are not covered by this one open set? Can you cover the rest of the set using finitely many more open sets?

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    Really great hint! Of course, $0$ is a cluster point. Thanks a lot! – 2017-01-25
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    @RaulGuarini you need $0$ is the limit of the sequence not just a cluster point. That is what gives you the "all but finitely many". – 2017-01-26
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Suppose you have some open cover ${\cal G}$, then there is some $U_0 \in {\cal G}$ such that $0 \in U_0$. Since $n \to {1 \over n}$ converges to $0$, there is some $N$ such that ${1 \over n} \in U_0$ for all $n \ge N$.

Now pick any $U_k \in {\cal G}$ such that ${1 \over k} \in U_k$ for $k=1,...,N-1$. Then $U_0, U_1,...,U_{N-1}$ is a finite cover.

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Let $U = \{U_{\alpha}\}$ be an open cover of your set. Each point in the set is in at least one $U_{\alpha}$ so for each $k$ in your set let $U_k$ be an open set containing $k$. So your set is a subset of $\{U_k|k \in the set\}$. And $V = \{U_k|k \in the set\}$ is a countable subcover.

Consider $0 \in U_0$. $U_0$ is open so there is an $\epsilon > 0$ so that $[0 \epsilon) \subset U_0$. Likewise we know for $n > 1/\epsilon$ then $0 < 1/n < \epsilon$ and $1/n \in [0,\epsilon) \subset U_0$. Furthermore all $1/m; m > n$ we have $1/m \in [0,\epsilon) \subset U_0$.

So $\{1, 1/2, ..... \frac 1{n-1} \} \subset \cup_{0< k < n} U_k$ and $\{0, 1/n, \frac 1{n+1},.....\} \subset U_0$.

So your set = $\{1, 1/2, ..... \frac 1{n-1} \}\cup \{0, 1/n, \frac 1{n+1},.....\} \subset \cup_{0< k < n} U_k \cup U_0 = \cup_{0 \le k < n} U_k$ and so $\cup_{0 \le k < n} U_k$ is a finite subcover.