Consider the space $X = \{ z \in \mathbb{C} : |z| \leq 1 \} \cup \{ z \in \mathbb{C} : |z| = 2 \}$ with the topology generated by the following basic open sets:
- the basic open neighbourhoods of $w \in X$ with $|w| < 1$ are just the usual (Euclidean) open neighbourhoods of $w$ in the open unit disc,
- the basic open neighbourhood of $w \in X$ with $|w| = 2$ is the singleton $\{ w \}$, and
- the basic open neighbourhoods of $w \in X$ with $|w| = 1$ are of the form $U \cup \{ 2z : z \in U, |z| = 1, z \neq w \}$ where $U \subseteq \{ z \in \mathbb{C} : |z| \leq 1 \}$ is open in the usual (Euclidean) topology on the closed unit disc.
Is this space $X$ compact?
The subspace $K = \{ z \in X : |z| = 1\text{ or }|z| = 2 \}$ is just the Alexandroff Double Circle, which is compact. I'm not sure if $X$ is similarly compact.