Smallest topology on $\mathbb{C}$ such that all singletons are closed This question was part of an exam for which I am preparing.

Question was related to finding smallest topology T on $\mathbb{C}$ in which all singletons are closed.
and then findind which options are true amog these:

A ($\mathbb{C}$,T) is haursdorff.
B ($\mathbb{C}$,T) is compact.
C ($\mathbb{C}$,T) is connected.
D  $\mathbb{Z}$ is dense in ($\mathbb{C}$,T) .
But  I am just a beginner in topology and self studying from Wayne patty . I was unable to find what topology should be . So, can you please tell what topology should be . rest I would prefer to work by mysely.
Thank you.
 A: I think the topology you are looking for is the co-finite topology, i.e.
$$ \tau= \{ A\subseteq X: \vert A^c\vert<\infty \} \cup \{ \emptyset\}.  $$
You just have to verify that every topology in which the singletons are closed, contains the co-finite topology, and that the co-finite topology is one of them.
A: I will answer the questions A and D, as you asked in the comments.
A is false because if we have two points and any the neighborhoods of them, say $U$ and $V$, they must satisfy $|\mathbb C\setminus U|,|\mathbb C\setminus V|<\infty$ and their intersection, $U \cap V$, is an open subset. If $U\cap V$ is empty, then its complement is union of finite sets, that is $\mathbb C= \mathbb C\setminus (U\cap V)=(\mathbb C \setminus U)\cap (\mathbb C\setminus V)$ is finite, an absurd!
For proving D, we take any non empty, open subset $U \subseteq \mathbb C$, we have $|\mathbb C\setminus U|<\infty$. If $\mathbb Z \cap U =\emptyset$, then $\mathbb Z \subseteq \mathbb C\setminus U$. That is, $\mathbb Z$ is finite, another absurd! So, $\mathbb Z \cap U \neq \emptyset$. Therefore, the integers are dense in $\mathbb C$ with that topology (in fact, any infinite set).
