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.
Can someone help me see the contradictions?
Thanks in advance!