We have a Cauchy sequence in the normed space of bounded functions $(f_n)_n \subset B(\Omega, \mathbb{K})$. I have shown that $(f_n(\omega))_n$ is a Cauchy squence for all $\omega \in \Omega$. What's the rigor behind being able to choose a natural number $N \in \mathbb{N}$ such that $$|f_n(\omega)- f_m(\omega)| \le \|f_n- f_m\|_{\infty} < \varepsilon \text{, for all } \omega\in \Omega \text{ and } n,m \ge N.$$
I mean, why exists $N = max_{\omega \in \Omega}\{N_{\omega}\}$ for $\Omega$ arbitrary, where $N_{\omega}$ is the corresponding Cauchy index for $(f_n(\omega))_n$? I have a common foundation in logic.
I don't understand what you mean with $N_\omega$. As I understood, the task is to show that such a $N_\omega$ exists.
So, let $\omega \in \Omega$ be arbitrary and $\varepsilon>0$. As the sequence $(f_n)_{n \in \mathbb{N}}$ is Cauchy (is suspect with the supremum norm, as it is the ''reasonable'' one on the space of bounded functions) there exists $N \in \mathbb{N}$ such that
$$ \|f_n-f_m\|_\infty<\varepsilon $$ for all $n,m\geq N$. As by definition of the supremum we have $|f_n(\omega)-f_m(\omega)|\leq\|f_n-f_m\|_\infty$ we get that also
$$ |f_n(\omega)-f_m(\omega)|<\varepsilon $$ holds for all $n,m\geq N$.
This shows that we can choose the same $N$ for all $\omega$.