A related question asks is there a smallest set that generates a given $\sigma$-algebra:
Smallest collection of subsets that generate a sigma algebra
The only existing answer to that (at the time of writing this) says that there must be a smallest cardinality for any generating set of any $\sigma$-algebra, by the well-orderedness of cardinals.
Focusing on the Borel sigma algebra over the Reals, $\mathcal{B}(\mathbb{R})$, the above implies that there's a smallest such cardinality of a set that generates this algebra. But is that smallest cardinality uncountable? I believe this is formally what I'm asking:
$$|\mathbb{N}| \notin \{|S| \mid S \in \mathcal{P}(\mathbb{R}) \wedge \sigma(S) = \mathcal{B}(\mathbb{R})\}$$
My intuition says yes. If there was a countable set generating the Borel algebra it would seem weird... but I'm not sure...
Open intervals $(a,b)$ with $a$ and $b$ rational form a countable class generating the Borel sigma algebra of $\mathbb R$.