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.
Following Willie Wong, the answer to his question depends on the cardinality of $X$.
In the case when $card(X)=c$ his question is not uniquely solvable
in the theory $(ZF)\&(DC)$ for $X=[0,1]$ and $\cal{F}=P[0,1]$, where
$(ZF)$ denotes the Zermelo-Fraenkel set theory and $(DC)$
denotes the axiom of Dependent Choices.
Indeed, on the one hand, in the consistent theory $ZF \&DC \& AD$, where $AD$ denotes an Axiom of Determinacy, the answer to his question is yes, because Mycielski and Swierczkowski well known result asserts that every subset of the real axis is Lebesgue measurable. Hence such a measure is exactly Lebesgue measure in $[0.1]$.
On the other hand, in the consistent theory $ZF\& DC \& AC \& \omega_1=2^{\omega}$ by Ulam's well known result on the powerset of $\omega_1$ (correspondingly, of $2^{\omega}$) we can not define a probability measure which vanishes on singletons.
Since both theories are consistent extensions of the theory $ZF\& DC$ we deduce that Willie Wong's question is not solvable within the theory $ZF\&DC$ for $X=[0,1]$ and $\cal{F}=P[0,1]$.
Best Answer
Why be fancy? A countable set can have uncountably many subsets. For example, take $\Omega=\mathbb N$ and $\mathcal A=2^\Omega$ and $P(X=n)=2^{-n}$.