This is one of the past paper problems to my topology finals, but Im not sure how to show this, I would appreciate the help it would really help me revise for my finals.
Problem: Suppose that $X$ is infinite. Then prove that $X$ is connected, compact but not Hausdorff with respect to the finite complement topology $\mathcal{T}_X.$ Where $X$ is the set with the finite complement topology.
To see that $X$ is connected, suppose it is not, then $X = X_1 \cup X_2$, where $X_1$ and $X_2$ are disjoint open sets. By definition of the finite complement topology, $X_1^c = X_2$ and $X_2^c = X_1$, so both are finite hence $X$ being a finite union of finite sets is finite, a contradiction.
To see that $X$ is compact, consider an open cover of $X$. Pick any open set $U$ in the open cover. By definition of finite complement, $U$ leaves out only finitely many points in $X$, and for each such point you can find an open set in the open cover that contains that point. Can you see why $X$ is compact now?
As for Hausdorffness, do you think that any two non-empty open sets in this topology can be disjoint? They can't, because then one is contained in the complement of the other, but every open set is infinite, so this would imply that an infinite set is contained in a finite set, a contradiction.
At this point, recall what Hausdorff is : for every pair of unequal points $x \neq y$ there exist disjoint open sets $x \in U_x, y \in U_y$.
So clearly, $X$ is not Hausdorff, since if we pick any two points, say $x,y$, then there are no disjoint pair of open sets containing $x$ and $y$ resp., since these open sets are non-empty, and above we explained why they have to intersect.