Infinite spaces in which all subsets are compact are not Hausdorff

Question

Let $(X,\tau)$ be an infinite topological space with the property that every subspace is compact. Prove that $(X,\tau)$ is not a Hausdorff space.

I start by supposing $X$ is Hausdorff. Then I can prove that every subspace of $X$ is normal. How can this lead me to a contradiction?

Answer 1

In a Hausdorff space, compact sets are closed. To see this, suppose $K$ is compact, and let $x\notin K$. Then for each $k\in K$, consider an open ball $U_k$ around $k$ and an open ball $V_k$ around $x$ which are disjoint. The collection of all such $U_k$ is a cover of $K$. Take a finite subcover, and the intersection of the corresponding $V_k$ is an open set which contains $x$ but does not intersect $K$.

From here it is not hard to describe the topology and find a noncompact set. Feel free to comment if you get stuck.

Answer 2

Assume $X$ is Hausdorff. If every subspace is compact then every subspace is closed. So the topology is discrete. Now take the cover by singletons. If this has a finite subcover then the space must be finite.