Prove that any family of disjoint 8-signs on the plane is countable

A good place to start is to ask yourself what the difference between the shapes of 8, T, and E on the one hand and M and O on the other. One thing that stands out is that each of 8, T, and E has a point that does not have a neighborhood that looks like a line segment: 8 has the point of tangency of the two circles, T has the intersection of the two straight lines, and E has the intersection of the vertical line and the middle horizontal line. (The other two horizontal lines of the E) can be straightened out so that it looks like $\vdash$. The M, on the other hand, can be straightened out into a line segment, and each point of the O looks like a point on C if you look at it up close, so it also looks like it’s on a line segment.

I would start by trying to show that a collection of pairwise disjoint T’s in the plane must be countable: they’re simpler than the 8, but every 8 contains a (slightly deformed) T, so if you can do it for T’s, you’ve basically done it for 8’s.

Suppose that $\mathscr{T}$ is an uncountable collection of pairwise disjoint T’s in the plane. For each $T\in\mathscr{T}$ there are rational numbers $p_T,q_T$, and $r_T$ with $r_T>0$ such that if $C_T$ is the circle of radius $r_T$ centred at $\langle p_T,q_T\rangle$, then the intersection point of $T$ is inside $C_T$, and the three endpoints of $T$ are outside $C_T$. There are only countably many triples of rational numbers, so there must be some $p,q,r\in\Bbb Q$ such that

$$\mathscr{T}_0=\{T\in\mathscr{T}:p_T=p,q_T=q,\text{ and }r_T=r\}$$ is uncountable.

Now trim the ends of each $T\in\mathscr{T}_0$ so that the three segments terminate exactly on the circle $C_T$; you now have a circle $C_T$ divided into three regions by the $T$ inside it. (If you bend the top bar of $T$ at the intersection point, this circle-with-T will look something like the Mercedes-Benz symbol.) Each of the regions must contain a point whose coordinates are both rational; call these points $x_T,y_T$, and $z_T$. There must be some $x,y,z\in\Bbb Q^2$ such that

$$\mathscr{T}_1=\{T\in\mathscr{T}_0:x_T=x,y_T=y,\text{ and }z_T=z\}$$

is uncountable. Can you show that if $T,T'\in\mathscr{T}_1$, then $T\cap T'\ne\varnothing$ and so get a contradiction? A sketch may help; note that $C_T=C_{T'}$.

I’ve added a suitable sketch below; $C=C_T=C_{T'}$, $P$ is the intersection point of $T'$, and $x,y$, and $z$ are as above.