Uncountably many 8s

There are many solutions to be found for this one. The easiest I know of is this...

Each loop of a figure- $8$ is open, and so must contain a point with rational coordinates.

If we have a collection $\mathcal{S}$ of disjoint figure- $8$ s, we can define a map $f \,:\, \mathcal{S} \to \mathbb{Q}^2 \times \mathbb{Q}^2$ , where $f(E) = (\mathbf{q}_1,\mathbf{q}_2)$ , where $\mathbf{q}_1$ is a point with rational coordinates inside one loop, and $\mathbf{q}_2$ is a point with rational coordinates inside the other loop of the figure- $8$ $E$ .

Suppose that $E_1,E_2 \in \mathcal{S}$ are distinct and such that $f(E_1) = f(E_2) = (\mathbf{q}_1,\mathbf{q}_2)$ . Since $E_1$ and $E_2$ are disjoint, the loop of $E_2$ that contains $\mathbf{q}_1$ is either totally inside the loop of $E_1$ that contains $\mathbf{q}_1$ , or else contains the whole of $E_1$ . Similarly, the loop of $E_2$ that contains $\mathbf{q}_2$ is either totally inside the loop of $E_2$ that contains $\mathbf{q}_2$ , or else contains the whole of $E_1$ . The only way in which these two loops can be mutually tangent is if they both contain the whole of $E_1$ . But, in this case, both $\mathbf{q}_1$ and $\mathbf{q}_2$ are inside both loops, which means that $E_2$ is not a figure- $8$ . This is impossible.

Thus we deduce that $f$ is injective, and hence that $\mathcal{S}$ is countable.