Prove that the disk $D(a;R)=\{z:\lvert z-a \rvert<R \} $ is not compact.
I know that we can prove that the set is not closed or not bounded and we can deduce directly that the set is not compact by the Heine-Borel Theorem, but I want to prove it directly from the definition.
Consider the metric space $(\mathbb{C}, \rho)$ whose matric is given by $$\rho: \mathbb{C} \times \mathbb{C} \to \mathbb{R}, \text{ with } \rho(z_1,z_2) = |z_1-z_2|.$$ Then is an open ball of radius $R$ centered at $a$, that is, $$D(a;R) := \{z \in \mathbb{C} : \rho(z,a) < R\}.$$ Now, a set is compact is every open cover has a finite subcover. For $n>1$, consider the set, $$\mathcal{O}_n := D(a; R-\frac{1}{n}).$$ Clearly, the collection $\{\mathcal{O}_n\}_{n\in \mathbb{N}}$ covers $D(a;R)$. Notice that for $z \in D(a; R)$ and sufficiently large $n$, $z \in D(a; R-\frac{1}{n})$ and as $n$ approach infinity, it does cover $D(a;R)$. But any finite subcover, say $\{\mathcal{O}_{n_j}\}_{j=1}^m,$ will cover only $D(a; R-\frac{1}{N})$, where $$N = \max \{n_j\}.$$ This proves that $D(a;R)$ is not compact.