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A set is countable if it admits an injective function into $\mathbb{N}$, equivalently if it is finite or can be listed in a sequence. The empty set is included; an infinite countable set has a bijection with $\mathbb{N}$.

Every rational number has a representation $p/q$ with $p\in \mathbb{Z}$ and $q\ge 1$. For each $k\ge 1$, there are only finitely many such pairs with $|p|+q=k$. List these finite levels successively, ordering each level by $p$, and discard repeated fractions when they first recur. Every rational number appears in some finite level, proving $\boxed{\mathbb{Q}\text{ is countable}}$.

For the uncountability proof, suppose all numbers in $(0,1)$ could be listed as $r_1,r_2,\dots$. Choose each decimal expansion not to end in an infinite string of nines, and denote its $j$th digit by $d_{ij}$. Set $b_i=1$ when $d_{ii}\ne 1$, and $b_i=2$ otherwise. Then the decimaldefines a number in $(0,1)$ with a unique such expansion, since its digits are only $1$ and $2$. It differs from $r_i$ at digit $i$ for every $i$, contradicting the enumeration. This Cantor diagonal argument proves $\boxed{\mathbb{R}\text{ is uncountable}}$ because its subset $(0,1)$ already is.

For the union of two countable sets $C,D$, take injective functions ${\iota }_C:C\to \mathbb{N}$ and ${\iota }_D:D\to \mathbb{N}$. The mapis injective: it is injective within each branch, and their images have different parity. Thus $\boxed{C\cup D\text{ is countable}}$, regardless of overlap.

The property of $A$ says that both $A$ and its complement are dense subsets of the real line. Both possible cardinalities occur. To establish the needed density directly, given $x\in \mathbb{R}$ and $\epsilon >0$, choose an integer $q>1/\epsilon$. Then $\lfloor qx\rfloor /q$ is rational and differs from $x$ by less than $1/q<\epsilon$. For an irrational approximation, choose a rational $r$ within $\epsilon /2$ of $x$ and a positive integer $q$ large enough that $\sqrt{2}/q<\epsilon /2$. The number $r+\sqrt{2}/q$ is irrational and lies within $\epsilon$ of $x$; irrationality follows from that of $\sqrt{2}$, since otherwise subtracting $r$ and multiplying by $q$ would make $\sqrt{2}$ rational. To recall the latter fact, a reduced fraction $p/q=\sqrt{2}$ would satisfy $p^2=2q^2$, forcing $p$ even and then $q$ even, contrary to coprimality. ThereforeThe second set is uncountable because, if it were countable, its union with $\mathbb{Q}$ would make $\mathbb{R}$ countable. The same two density arguments give the required approximations from $A$ and from its complement in either example.

Finally, the condition on $B$ makes every one of its points an isolated point, so $B$ is a discrete subset of the real line. Enumerate all rational intervals $(p_j,q_j)$ with $p_j<q_j$; enumerate pairs of indices of the rational list by increasing sum and keep those whose endpoints have the required order. For each $b\in B$, its isolating neighbourhood contains a rational interval $(p_j,q_j)$ containing $b$ and no other point of $B$. Assign to $b$ the least such index $j$. Two distinct points cannot receive the same index, because that interval would then contain two members of $B$. This gives an injective function $B\to \mathbb{N}$ and proves that discrete subsets of the real line are countable:Using the least index makes the assignment explicit and requires no arbitrary selection of an uncountable family of intervals.