Let $(\Omega,\Sigma)$ be a measurable space. An atom of $\Sigma$ is a set $B\in\Sigma$ such that for all $A\subseteq B$ either $A=\emptyset$ or $B=A$. A measurable space is atomic if every element lies in some atom. The $\sigma$-algebra $\Sigma$ is countably generated if there is a countable family of measurable sets such that $\Sigma$ is the smallest $\sigma$-algebra containing all of them. For example $(\mathbb{R},\mathcal{B})$ is countably generated since $\mathcal{B}$ is generated by the open intervals with rational endpoints. The atoms of $\mathcal{B}$ are the singletons.
Proposition: If $\Sigma$ is countably generated, then $(X,\Sigma)$ is atomic.
Proof: If there is a countable family generating $\Sigma$, there is also a countable family closed under complementation that generates $\Sigma$. If $\mathcal{C}$ is such a family, we get all atoms of $\Sigma$ as the intersection of all elements of $\mathcal{C}$ that contain a given point.
Now if $(\Omega,\Sigma,\mu)$ is a probability space, we call $B\in\Sigma$ a $\mu$-atom if $\mu(B)>0$ and for all $A\in\Sigma$ such that $A\subseteq B$, either $\mu(A)=0$ or $\mu(A)=\mu(B)$. The probability space is atomless if it contains no $\mu$-atom.
Lemma: If $(\Omega,\Sigma,\mu)$ is a probability space such that $\Sigma$ is countably generated and $\mu$ takes on only the values $0$ and $1$, then there exists an atom $A\in\Sigma$ such that $\mu(A)=1$.
Proof: Let $\mathcal{C}$ be a countable family closed under complementation that generates $\Sigma$. For each element of $\mathcal{C}$, either itself or its complement has probability one $1$. The intersection of all elements in $\Sigma$ with probability $1$ is an atom with probability $1$.
Proposition: If $(\Omega,\Sigma,\mu)$ is a probability space with $\Sigma$ countably generated, then it is atomless if and only if every atom in $\Sigma$ has probability $0$.
Proof: Clearly, in an atomless probability space, every atom must have probability $0$. Supppose now that $A$ is a $\mu$-atom. Let $A\cap\Sigma=\{A\cap S:S\in\Sigma\}$ be the trace $\sigma$-algebra. It is countably generated too. Then $(A,A\cap\Sigma,1/\mu(A)\cdot\mu)$ is a probability space such that the probability takes on only the values $0$ and $1$. So by the lemma, there is an atom $B$ such that $1/\mu(A)⋅\mu(B)=1$. But $B$ is also an atom of $\Sigma$ and $\mu(B)>0$.
So it follows that a probability measure on $(\mathbb{R},\mathcal{B})$ is atomless if and only if it puts probability $0$ on all singletons, which justifies the definition in the book of Kai Lai Chung.
Finally, an example of a probability space in which each atom has probability $0$ but such that the space is not atomless. Let $\Omega$ be any uncountable set, let $\Sigma$ consists of those subsets of $\Omega$ that are either countable or have an uncountable complement. Let $\mu(A)=0$ if $A$ is countable and $\mu(A)=1$ if its complement is countable. Every set with countable complement is an $\mu$-atom, but the atoms of $\Sigma$ are the singletons which all have probability $0$. Note that $\Sigma$ is not countably generated.
For the first question:
Let $A = \bigcup_{n \in \mathbb N} \{a_n \} \subset \mathbb R$. The Lebesgue measure of a point is zero: by construction of the Lebesgue measure, $\lambda [a,b] = b - a$. For the one element set $\{ a \} = [a,a]$ we have $\lambda [a,a] = a-a =0$.
Since the Lebesgue measure is additive, we have
$$ \lambda A = \sum \lambda \{a_n\} = \sum 0 = 0$$
Best Answer
If $\mu (A) =0$ or $\infty$ for all $A$ then the result is trivial. If that is not the case then we can restrict our attention to a set $A$ with $0 <\mu (A) <\infty$, so we may assume that $\mu$ is a finite measure.
Observe that there is no set $E$ with $0 <\mu (E) <a$. (1)
Suppose there is no atom. Then there exists $A_1\subseteq A$ such that $0<\mu(A_1)<\mu(A)$. By induction there exists a seqeunce $(A_n)$ such that $A_{n+1}\subseteq A_n$ and $0<\mu(A_{n+1})<\mu(A_n)$. Now $(A_n\setminus A_{n+1})$ is a disjoint sequence so $\sum \mu (A_n\setminus A_{n+1})\leq \mu (A)<\infty$. But then $\mu (A_n\setminus A_{n+1}) \to 0$ as $ n \to \infty$. But (1) shows that $\mu (A_n\setminus A_{n+1})$ must be $0$ for $n$ sufficiently large. Since $A_{n+1}\subseteq A_n$ this gives $\mu (A_{n+1})=\mu (A_n)$ contrary to our construction of $(A_n)$.