I'm studying about invariant cardinals and theirs natural generalizations to uncountable cardinals. I was reading a paper which focus over splitting number $\mathfrak{s}(\kappa)$ and there I found the following theorem:
Theorem: Let $\kappa$ be a uncountable cardinal. $\mathfrak{s}(\kappa)\geq\kappa$ iff $\kappa$ is a strongly inaccesible cardinal.
In particular, I'm interesting in the case when $\kappa=\aleph_1$ and it is clear that the theorem above implies that $\mathfrak{s}(\aleph_1)=\aleph_0$. But I want to get a direct proof of this fact however, I have troubles to achieve it.
I don't want a solution of my problem, I only need some suggest or advice to see my problem since another perspective. Thanks for advance.
Hint: Consider an $\aleph_1$ sized subset $X$ of $2^\omega$ and the sets $\{x \in X: s \subseteq x \}$ for $s \in 2^{<\omega}$.