Counter Example of $\cup$ A$_i$ is not compact. True or false: The union of an arbitrary family of compact subsets is a compact subset. Well firstly I think for a while if it is true or not, definitely I take the second option and in order to prove that it false I think for the next counter example: I take $\mathbb{R}$ as the working space with the discrete topology, so I think about the subset of only one element that all of them are compact so as $\mathbb{R}$ is non numerable we can't affirm that the union of all of this subset is compact because is a non numerable union so if we try to obtain a finite union that contains all of the elements it will be impossible. I'm not sure if it is right or not.

In $\Bbb R$ in the discrete metric/topology, a subset $A$ is compact iff $A$ is finite. So e.g. take $A_n = \\{n\\}$, all compact, and then $\Bbb N = \bigcup_n A_n$ is not compact. So it already fails with countable unions of finite sets.