Prove the set $K = \{\frac{1}{n} \mid n\in \mathbb{N}\}\cup \{0\}$ is compact.
I have tried to write a proof for this fact. Please critique it.
Proof.
Let $\mathcal{G} = \{G_\alpha \mid \alpha \in A\}$ be any open cover for $K$. Since $\mathcal{G}$ is an open cover, we have $K\subseteq \bigcup_{\alpha\in A}{G_\alpha}$. Hence, $0\in K$ implies that there exists some $G_0 \in \mathcal{G}$ such that $0\in G_0$. Since $G_0$ is open, there exists $r>0$ such that $B_r(0) \subseteq G_0$. We know from previous results that $\frac{1}{n} \to 0$. Hence, $B_r(0)$ contains all but finitely many terms of $\frac{1}{n}$ and consequently, $G_0$ contains all but finitely many of these points. Let $N$ be such that $\frac{1}{n} \in G_0$ for all $n\geq N$. Then for each point $1,\frac{1}{2},\dots,\frac{1}{N-1}$ we can choose a corresponding open set $G_1,G_2,\dots,G_{N-1} \in \mathcal{G}$ that contains each point. Hence, $\{G_0,G_1,G_2,\dots\,G_{N-1}\}$ is a finite subcover for $K$.