TITLE: in a topological space only finite subsets are compact sets
QUESTION [11 upvotes]: If in a topological space only finite subsets are compact sets, is it then the discrete topological space? 
Thank you.

REPLY [10 votes]: The examples given by bof are quite nice.
On the other hand, for $T_1$ and first countable spaces, being "anticompact" is equivalent to being discrete. (If you require compact spaces to be Hausdorff, as Bourbaki does, then "$T_1$" should be replaced by "Hausdorff").
Indeed, assume that $X$ is a $T_1$, first-countable and anticompact space. To show that $X$ is discrete, it is enough to show that any point $x\in X$ has a finite neighbourhood (then, since $X$ is $T_1$,  it follows that in fact every point is isolated).
Assume that some point $x\in X$ has the property that every neighbourhood of $x$ is infinite. Let $(V_n)_{n\in\mathbb N}$ be a countable basis of neighbourhoods for $x$. Since each $V_n$ is infinite, one can construct a sequence $(x_n)\subset X$ such that the $x_n$ are pairwise distinct and $x_n\in V_n$ for all $n$. Then the sequence $(x_n)$ converges to $x$, so $K=\{ x\}\cup\{ x_n;\;n\in\mathbb N\}$ is compact. This contradicts the anticompactness of $X$.