This is the proof I wrote for $D_1(0)=\{z\in \mathbb{C} \mid |z|< 1\}$ is not compact.
$$ \bigcup_{n=2}^{\infty}D_{1-(1/n)}(0) $$ is clearly a open covering of $D_1(0)$.
Consider the finite sub-collection $\bigcup_{n=2}^{k}D_{1-(1/n)}(0)$ of $\bigcup_{n=2}^{\infty}D_{1-(1/n)}(0)$.
Assume that $D_1(0) ⊆ \bigcup_{n=2}^{k}D_{1-(1/n)}(0)$. Choose $z\in \mathbb{C}$ such that $|z|=1-(1/2k)$.
Then $z\in D_1(0)$ but $|z|=1-(1/2k)>1-(1/k)$. Hence $z∉\bigcup_{n=2}^{k}D_{1-(1/n)}(0)$. This is a contradiction hence $D_1(0)$ is not compact.
I am not sure whether this proof is correct. Hope someone could help to improve this. THanks
Compact subsets of Hausdorff spaces are closed. Alternatively, the continuous function $z\mapsto (1-z)^{-1}$ on this set is not uniformly continuous, alternatively it doesn't achieve a maximum.