Let $X$ be an infinite set with a topology $T$, such that every infinite subset of $X$ is closed. Prove that $T$ is the discrete topology. Let $X$ be an infinite set with a topology $T$, such that every infinite subset of $X$ is closed. Prove that $T$ is the discrete topology.


I have somewhat of an answer but I don't think it's enough to prove it, especially with respect to the subsets being infinite.

Let $S$ be contained in $X$, then $X \setminus S$ is also contained in $X$. Therefore we can say that $X \setminus S$ is closed, therefore $S$ is open for any $S$ contained in $X$. Hence $T$ is the discrete topology.

Thanks
If $X$ is infinite, let $x \in X$ ; it follows that since $\{x\}$ is finite, $X \backslash \{x\}$ is infinite, hence closed. Therefore, for any subset $S \subseteq X$, 
$$
S = \bigcup_{x \in S} \{x\}
$$
is a union of open sets, thus open. This means $T$ is all subsets of $X$. You have the discrete topology.

Hope that helps,
Every infinite subset of X closed means every finite subset of X is open. In particular, the singleton sets {x} for all x in X are open. This is sufficient to conclude that T is discrete!

Hope it helps.