Prove that $K$ is a compact set by directly using the definition of compactness. Let $X = \mathbb R$ with usual distance metric. Let $K = \{0\}\cup \{1/n | n \in N\}$. Prove
that $K$ is a compact set by directly using the definition of compactness.
Definition: A set $C \subset E$ is said to be compact if every collection of open sets that covers $C$ has a finite sub-collection that covers $C$. The metric space $(E,d)$ is said to be compact if E is so.
Can we start by constructing balls centered at $1/n$ with radius 1 and say it is an open cover of K. But since in the definition it says every open collection I do not know if it right to select a specific one.
 A: Let $\mathscr{U}$ be a cover of $K$; then, for each $n>0$, there is $U_n\in\mathscr{U}$ such that $1/n\in U_n$.
There is also $U_0\in\mathscr{U}$ such that $0\in U_0$.
Can you go on?
A: Right: you cannot prove this starting from a specific cover.
If $\mathcal U$ is an open cover of $K$, there is some $A_0\in\mathcal U$ such that $0\in A_0$. Since $A_0$ is open and $\lim_{n\to\infty}\frac1n=0$, $\frac1n\in A_0$ is $n$ is large enough. So, there is some $N\in\mathbb N$ such that $n\geqslant N\implies\frac1n\in A_0$. For each $n\in\{1,2,\ldots,N-1\}$, let $A_n\in\mathcal U$ be such that $\frac1n\in A_n$. Then $\bigl\{A_n\,|\,n\in\{0,1,\ldots,N-1\}\bigr\}$ is a finite subcover of $\mathcal U$.
A: Hint:Let $\{U_{\alpha}\}$ be an open cover of $K$. Let $0\in U_{\alpha_0}$. Since $U_{\alpha_0}$ is open, there exists  $r\gt 0$ such that $B(0,r)\subseteq U_{\alpha_0}$. As $0$ is the limit point of the set$K\setminus \{0\}= \{1/n | n \in \Bbb N\}$,there are only finitely many points of $K\setminus\{0\}$ outside of $B(0,r)$. Can you proceed from here? 
