Consider $\mathbb{Z}$ equipped with a topology generated by the basis of the form $(-k,k)$, is a finite set, say (-3,3), compact? Is $\mathbb{Z}$ compact?
I think a finite set is compact since you can just find a finite (in fact singleton) subcover, that is the interval that includes all the points in the finite set. For the second part, $\mathbb{Z}$ is not for the same reason that it is not compact in the standard topology.
Can someone please check if what I am thinking is correct? Thanks
Let $(X, \tau)$ be a topological space. Suppose that $A \subseteq X$ so that $|A|<\infty$. Suppose that $\mathcal{U}$ is an open cover of $A$. Then we go ahead and choose $U_a$ so that $a \in U_a$. There are at most finitely many $U_a$. Then $A^{\prime}=\{U_a \mid a \in U_a\}$. But then there is a finite sub-collection so that $A \subseteq A^{\prime}$, hence $A$ compact. This is more general. Any finite set is compact.
I'm not sure what you meant by "singleton sub-cover," but the set of $\{x\} \subseteq X$ need not be open. See here for further discussion: Are Singleton sets in $\mathbb{R}$ both closed and open?