There is something confusing about the terminology "discrete". Discrete implies some sort of topology, some sort of sense to say whether two elements are close to one another. Discrete means that the elements are spaced.
If we talk about subsets of the real numbers, then finite sets are always discrete; and every discrete set is countable (or finite). However the rationals are not a discrete set, between two rationals you can always find another rational.
When I was a freshman we always assumed that discrete is interchangeable with countable (or finite), and I learned only later that this is a flawed concept. Discrete sets can be uncountable, in the broad context of mathematics, and finite sets can be made non-discrete as well in the broad context of mathematics.
What I do read from your question is whether or not countable includes finite. This depends on the context, and whether or not it would simplify things for us.
Sometimes we want to say that finite is countable, because it means things easier "A set is countable if and only if it is equipotent to a subset of the natural numbers" would include finite in the definition.
On the other hands, there are cases where the finite cases are irrelevant and serve as a point which we have to deal with separately, so defining countable to be countably infinite makes things easier. You deal with the finite case, and then you deal with the countable case.
In either case it is common in textbooks and papers to include a word on the meaning of countable. We could mean "at most countable" or "countably infinite". Usually one can deduce that from the context, if the other term appears.
That is, if we run into something like this:
If $A$ is countable, and $f\colon A\to\mathbb R$ is this and that, then the image of $f$ is at most countable.
would imply that countable means countably infinite, however is not necessarily the case and one should probably look for the definition in the preliminaries section as well.
Suppose that $\lvert \Omega\rvert\ge\aleph_0$, and $\mathscr M\subset\mathscr P(\Omega)$ is a $\sigma-$algebra. We shall show that:
$$
\textit{Either}\,\,\,\, \lvert\mathscr M\rvert<\aleph_0\quad or\quad \lvert\mathscr M\rvert\ge 2^{\aleph_0}.
$$
Define in $\Omega$ the following relation:
$$
a\sim b\qquad\text{iff}\qquad
\forall E\in\mathscr M\, (\,a\in E\Longleftrightarrow b\in E\,).
$$
Clearly, "$\sim$" is an equivalence relation in $\Omega$, and every $E\in\mathscr M$ is a union of equivalence classes. Also, for every two different classes $[a]$ and $[b]$, there are $E,F\in\mathscr M$, with $E\cap F=\varnothing$, such that $[a]\subset E$ and $[b]\subset F$.
Case I. If there are finitely many classes, say $n$, then each class belongs to $\mathscr M$, and clearly $\lvert \mathscr M\rvert=2^n$.
Case II. Assume there are $\aleph_0$ classes. Fix a class $[a]$, and let
$\{[a_n]:n\in\mathbb N\}$ be the remaining classes. For every $n\in\mathbb N$, there exist $E_n,F_n\mathscr\in M$, such that $[a]\subset E_n$, $[a_n]\subset F_n$ and $E_n\cap F_n=\varnothing$. Clearly, $[a]=\bigcap_{n\in\mathbb N} E_n\in\mathscr M$, and thus $\lvert \mathscr M\rvert=2^{\aleph_0}$.
Case III. If there are uncountably many classes,
we can pick infinite countable of them $[a_n]$, $n\in\mathbb N$, and disjoint sets $E_n\in\mathscr M$, with $[a_n]\subset E_n$, (using the Axiom of Choice), and then realise that the $\sigma-$algebra generated by the $E_n$'s has the cardinality of the continuum and is a subalgebra of $\mathscr M$.
Best Answer
The set of all subsets of $\mathbb N$, also called the powerset of $\mathbb N$. It is denoted by $P(\mathbb N)$ and has a cardinality of $\beth_1=2^{\aleph_0}$ which is uncountably infinite.