The smallest topology in $\mathbb C$ in which every singleton set is closed 
Let $\tau$  be the smallest topology on $\mathbb C$  such that every singleton set under $\tau$ is closed. Then which of the following is true ?
$1.(\mathbb C,\tau) \text{ is  not Hausdorff}.$
$2.(\mathbb C,\tau) \text{ compact }$
$3.(\mathbb C,\tau) \text{ connected }.$
$4. \mathbb Z \text{ is dense in } (\mathbb C,\tau).$

All $4$ options are  correct if I consider this to be the co-finite topology on the Complex numbers.  But I do not know what the smallest topology would be. It won't be the discrete one but what other options do we have here ?
Please discuss.Thank you.

First of all, notice that nothing about this problem is particular to $\Bbb C$; you can replace it with any infinite set whatsoever (and then $\Bbb Z$ is replaced by your favorite infinite subset).
It does turn out to be true that the smallest suitable topology is the co-finite topology (so awesome that you know the answers for that topology already!). One needs to show that any topology in which every singleton set is closed contains the co-finite topology. So let $\tau$ be such a topology. Can you show that any set $U$ that is open in the co-finite topology must also be open in $\tau$, just using the properties of topologies (such as the finite intersection property) and the one extra thing we know about $\tau$?
