My textbook says that if $X: \Omega \to \mathbb{R}$ is discrete stochast (I.e., there are only countably many values that get reached), then it suffices to know the probability function $p(x) = \mathbb{P}\{X =x\}$ in order to know the distribution function $$\mathbb{P}_X: \mathcal{R} \to \mathbb{R}: A \mapsto \mathbb{P}\{X \in A\} = \mathbb{P}(X^{-1}A)$$
Indeed, if $S:= \{x : p(x) > 0\}$, then for $A \in \mathcal{R}$, it follows that $$\mathbb{P}\{X \in A\} = \sum_{x \in S \cap A}p(x)$$
But how do they get this formula?
I tried the following:
$$\mathbb{P}\{X \in A\} = \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S\cap A}\{a\} \cup\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$= \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S \cap A}\{a\}\right)\right) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$ $$=\sum_{a \in S \cap A}p(a) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right)$$
But how do I show that the probability on the right is zero? I can't use $ \sigma$-additivity on uncountable disjoint unions.
To prove this from a measure theoretic perspective you need first to clarify what the measure space $(\Omega, \mathcal F, \mathbb P)$ is.
Assuming $\Omega$ is discrete
Since we consider a discrete distribution, without loss of generality we can assume it is a distribution on $\mathbb N$, so that we can choose to use $\Omega = \mathbb N$.
The cannonical choice of $\sigma$-algebra is then $\mathcal F = 2^{\mathbb N}$ the power set of $\mathbb N$; moreover, note that this is also the $\sigma$-algebra generated by $\mathbb N$. This means that it suffices to define the measure $\mathbb P$ on the individual integers, i.e. $\mathbb P$ is determined by $\mathbb P(\{n\}), \, n \in \mathbb N$.
In particular we note that any measurable $A \in \mathcal F$ must be countable, as it is a subset of $\mathbb N$ which is itself countable. From here, your proof above is trivial because countable sub-additivity is a given.
Assuming $\Omega = \mathbb R$
In this case we suppose that the $\sigma$-algebra chosen is the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$, which is generated by intervals $(a,b)$ and define
$$\mathbb P\big( (a,b) \big) = \sum_{ x \in S \cap (a,b) } p_x,$$ note that this is how we define the measure on the measurable sets: it is not a property of it.
In conclusion, it is impossible to answer your question clearly without knowing what assumptions were made on the measure space.
In the first of the above we assume the space to be countable from which the property follows from countable additivity.
In the second we assume it is continuous but then have to define an appropriate $\sigma$-algebra and probability measure, but as done above if defined in the canonical fashion this makes the property trivial.