$\text{cf}(\aleph_{\omega_1})=\omega_1$
From here, I quote Brian M. Scott's proof:
Suppose that $\langle\alpha_n:n\in\omega\rangle$ is an increasing sequence cofinal in $\omega_{\omega_1}$. For each $\xi<\omega_1$ there is a least $n(\xi)\in\omega$ such that $\omega_\xi\le\alpha_{n(\xi)}$. Clearly there is then an $m\in\omega$ such $X=\{\xi<\omega_1:n(\xi)=m\}$ is uncountable, and it follows easily that $\color{blue}{\omega_{\omega_1}=\sup_{\xi\in X}\omega_\xi}\le\alpha_m<\omega_{\omega_1}$, which is absurd.
I would like to ask what is the reasoning behind $\color{blue}{\omega_{\omega_1}=\sup_{\xi\in X}\omega_\xi}$.
I think we can prove $X$ is cofinal in $\omega_1$ from the fact that $X\subseteq \omega_1$ and $|X|=\omega_1$. I have tried but unsuccessfully.
Thank you for your help!
Every initial segment of $\omega_1$ is countable, so $X$ is unbounded in $\omega_1$
Like Asaf commented, the fact that there exists such $X$ require choice, without some kind of choice it is not necessarily correct