I just started studying Topology this week so bear with me if the answer is trivial...
Given a topology $(X, \tau)$, where $X$ is an infinite set. Show that if all the infinite subsets of $X$ are closed, then $(X, \tau)$ is the discrete topology.
My approach is the following. Suppose that $S \subset X$. Then either $S$ is finite or infinite.
If $S$ is finite, then $X \setminus S$ is infinite (because $X$ infinite), and because $X \setminus S$ is closed, $S$ is open.
If $S$ is infinite, then I run into troubles. You see, take the natural numbers $X = \mathbb N$. Then $S = \mathbb N \setminus \{1 \}$ is infinite and its complement is finite. So there is no guarantee that $X \setminus S$ is infinite right? Do I miss something here?
You can see that if $x\in X$ is a point then $\{x\}$ is open, so every singleton is open and hence our space is discrete.