Is the Discrete Topology on $X$ the Only One Containing All Infinite Subsets of $X$? Prove or find counterexamples.
Let $X$ be an infinite set and $T$ be a topology on $X$. If $T$ contains every infinite subset of $X$, then $T$ is the discrete topology.
 A: Suppose $T$ is a topology containing all the infinite subsets of $X$. I claim every finite subset also belongs to $T$, and so $T$ is the discrete topology.
To see this, let $A$ be any finite subset of $X$. Since $X$ is infinite, $X \setminus A$ is infinite. Partition $X \setminus A$ into two disjoint infinite subsets $Y_1$ and $Y_2$ (this can always be done if the Axiom of Choice is assumed).
Now, $Y_1 \cup A$ and $Y_2 \cup A$ are both infinite sets, so they belong to $T$. Moreover, their intersection is precisely $A$. Since topologies are closed under finite intersection, it must be the case that $A$ belongs to $T$. Since $A$ was an arbitrary finite set, the claim follows.
A: Let's suppose that $X$ is an amorphous set, in the cofinite topology. By the amorphous nature of $X$, every infinite subset of $X$ has a finite complement, so every infinite subset of $X$ is open. Thus, the open subsets of $X$ are precisely the empty set, $X$, and the infinite proper subsets of $X$. However, this is not discrete, as (for example) no singleton subset of $X$ is open.
Now, if we have enough Choice so that there aren't any amorphous sets, then Austin's approach is the way to go for a proof. Otherwise, the above serves as a counterexample.
Remark: I don't intend this to be a "competing" answer with Austin's. I intended merely to elucidate why I brought up the Axiom of choice and why he then felt compelled to make mention of it in his answer. He answered before I did, it's a good answer, and you didn't pre-specify how much (if any) Choice you're using. If you like mine, feel free to upvote, but if you're debating which of our answers to accept, go with his.
