Every compact set $S\in \mathbb{C}$ is bounded

Question

This is my proof for every compact set $S \subseteq \mathbb{C}$ is bounded.

Let $S \subseteq \mathbb{C}$ be compact and assume that it is not bounded.

Then for each $z\in \mathbb{C}$ and for each $r>0$, $S\not\subset D_r(z)$.

Let $z\in \mathbb{C}$. Then $\bigcup_{n=1}^{\infty}D_{n}(z)$ is an open covering of $S$. Also $\big\{D_{n_i}(z)\,\big|\, i \in \{1,2,...,k\}\big\}$ is a finite sub collection of $\bigcup_{n=1}^{\infty}D_{n}(z)$. Then $S\subset \bigcup_{i=1}^{k}D_{n_i}(z)$.

But $\bigcup_{i=1}^{k}D_{n_i}(z)=D_{n_k}(z)$ and $S \not\subset D_{n_k}(z)$ since it is not bounded. This is a contradiction. Thus, $S$ is compact.

I hope someone could verify this proof or suggest something better. Thanks

Answer 1

This proof seems perfectly fine. Well done!

The only (stylistic) change I would make is that since there is no reason for $z$ to be arbitrary, consider $D_n(0)$ rather than $D_n(z)$ for some unspecified $z \in \Bbb C$.