For context, I am self studying some basic topology in preparation for taking Algebraic Topology (following Hatcher), and I've come across the following practice problem. The source is Math 327 (2019 summer) from University of Toronto (4.18 in the Big List). I have no prior experience with topology (except for whatever I vaguely remember from analysis).
Call a subset $X \subseteq \mathbb{R}^2$ a “$Y$-set” if it is the union of three straight line segments that share a common endpoint. (That is, a set that looks like a capital Y, except the three lines can meet at any angles and can each be of any finite length.) Prove that any collection of mutually disjoint $Y$-sets is countable.
I do not see how this can be true. For instance, note that there is no restriction on the angles at which the lines meet (in fact it emphasizes this). In particular, any line segment is a $Y$ set, and certainly the set of all vertical line segments with one base point on the line $y=0$ and the other on the line $y=1$ is not countable (since obviously the reals are not countable).
Now, however, if we amend this question to specify that the lines cannot be colinear (so it looks like a Y, sort of). Of course the construction I've made no longer works and I cannot think of a simple way to make it work but I suspect there exists a similar counterexample.
I've also noticed that if each set has a rational point then we have an injection into $\mathbb Q^2$ since each $Y$ is disjoint and thus it's countable. Also if there exists disjoint open covers of each $Y$ then we are also done by the fact that $\mathbb R^2$ satisfies the countable chain condition. So if we want a counterexample, it really has to be some strange dense set of $Y$s.
Is this something particular about being colinear that I'm not seeing here? Or is this question unfortunately incorrect?