= Solution
A <countable set> is a set that is finite or can be put in <bijection> with the <natural numbers>; equivalently, it admits an <injective function> into the natural numbers. To enumerate the <rational numbers>, list pairs $(p,q)$ with $p\in\mathbb Z$, $q\ge1$ in increasing order of $|p|+q$, and then list $p/q$, discarding repetitions. Each level is finite and every rational eventually appears. An enumeration $q_1,q_2,\ldots$ of the rationals then enumerates $\mathbb Q\times\mathbb Q$ by listing $(q_i,q_j)$ in diagonals of constant $i+j$. Thus
$$
\boxed{\mathbb Q\times\mathbb Q\text{ is countable}.}
$$
To prove that the <real numbers> are <uncountable>, suppose $(0,1)$ had an enumeration $x_1,x_2,\ldots$. Write each $x_n$ in its decimal expansion that does not end in an infinite string of nines. Define a new decimal $y=0.d_1d_2\cdots$ by taking $d_n=2$ if the $n$th digit of $x_n$ is $1$, and $d_n=1$ otherwise. Then $y\in(0,1)$, has no ambiguous terminating/nines representation, and differs from $x_n$ at digit $n$. The <Cantor diagonal argument> contradicts the supposed enumeration. Consequently \b[the real numbers are uncountable].
For the discs, use the enumeration of $\mathbb Q^2$ just constructed. Every positive-radius disc contains a rational-coordinate point: its interior contains a small open rectangle, and each coordinate interval contains a rational number. For each disc $D$, let $j(D)$ be the least index of an enumerated rational-coordinate point in its interior. This minimum exists and requires no arbitrary simultaneous choice. Disjoint discs cannot receive the same point, so $D\mapsto j(D)$ is an <injective function> into the natural numbers. This proves the <countability of pairwise disjoint open disks> and also the same conclusion for closed discs, by using their disjoint interiors.
For circles, take all concentric <circles> centred at the origin with radii $r\in(1,2)$. Two different radii give disjoint <circles>, and every radius is positive. The radius map is a <bijection> from the uncountable interval $(1,2)$ to this family. \b[An uncountable disjoint family of nontrivial circles is therefore possible:] the rational-point argument for discs relies on their nonempty interiors, which circumferences do not have.
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