Call a tack the one point union of three open intervals. Can you fit an uncountable number of them on the plane? Or is only a countable number?

First of all, the onepoint compactification of three open intervals is not a "tack", it's a threeleaf clover. I think that you mean a onepoint union of three closed intervals; of course it doesn't matter if the other three endpoints are there or not. This topological type can be called a "Y" or a "T" or a "simple triod". R.L. Moore published a solution to your question in 1928. The answer is no. It was generalized in 1944 by his student Gail Young: You can only have countably many $(n1)$dimensional tacks in $\mathbb{R}^n$ for any $n \ge 2$. For her theorem, the name "tack" makes rather more sense, but she calls it a "$T_n$set". Actually Moore's theorem applies to a more general kind of triod, in which three tips of the "Y" are connected to the center by "irreducible continua", rather than necessarily intervals. I don't know whether I might be spoiling a good question, but here in any case is a solution to the original question (see as both Moore and Young did something more general that takes more discussion). Following domotorp's hint, there is a Suppose that you have uncountably many simple triods in the plane. Given a simple triod, we can choose a circle $C$ with rational radius and rational center with the branch point of the triod on the inside and the three tips on the outside. Since there are only countably many such circles, there are uncountably many triods with the same circle $C$. We can trim the segments of each such triod so that they stop when they first touch $C$, to make a pie with three slices (a MercedesBenz symbol). Then, given such a triod, we can pick a rational point in each of three slices of the pie. Since there are only countably many such triples of points, there must be uncountably many triods with the same three points $p$, $q$, and $r$. In particular there are two such triods, and a suitable version of the Jordan curve theorem implies that they intersect. The argument can be simplified to just pick a rational triangle that functions as the circle, and whose corners function as the three separated points. But I think that there is something to learn from the variations together, namely that the infinite pigeonhole principle gives you a lot of control. For instance, with hardly any creativity, you can assume that the triods are all large. 


This is a wellknown puzzle/problem, the trick is to make an injective mapping from any set of disjoint tacks to triples of $\mathbb Q^2$. 

