Contradiction in the proof a space with every compact subspace is 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.

A space is Husdorff if for all $a,b\in X$ then there exists two open sets $V,U\in\tau$ such that $a\in U,b\in V$ and $U\cap V=\emptyset$.

Since all the subspaces are compact means all the elements of X are closed and then opened at the same time by the compliment so we are dealing with the discrete topology.

If I assumed the space was Hausdorff and all subspaces were compact I believe I would get a contradiction, but I do not see how.

Question:

Can someone help me see the contradictions?

Thanks in advance!

Answer 1

Hint: if $X$ is Hausdorff, then every subspace of $X$ is compact thus closed. What can you say about $\tau$ then?

Answer 2

• If $X$ were Hausdorff then $(X,\tau)$ is discrete (as you rightly noted).
• If $X$ is discrete and compact (as it must be, as all subspaces are) it is finite. (Consider the open cover by singletons).
• Contradiction, as $X$ is supposed to be infinite.