The question I am interested in is the following:
Let $\omega_1$ be the first ordinal with an uncountable number of predecessors. We consider $X=[0,\omega_1]$ supplied with order topology. Prove that if $\{K_n\}_{n\in\mathbb{N}}$ is a sequence of uncountable compact sets of $X$, then $\bigcap\limits_{n\in\mathbb{N}}K_n$ is uncountable.
This suggests to me that an uncountable compact $K$ contains necessarily a subset of $X$ of the form $[\alpha,\omega_1]$: this result would indeed imply that the intersection of the sets $K_n$ contains such an uncountable set since $\omega_1$ cannot be the limit of a sequence of points in $[0,\omega_1[$. I can see that any $K_n$ contains $\omega_1$, otherwise $\cup_{\alpha<\omega_1}[0,\alpha]$ is an open cover of $K_n$ from which we cannot extract a finite subcover. But I can't figure out a way to prove that $K_n$ contains a set of the form $[\alpha,\omega_1]$. Is it true?
The set of limit ordinals in $[0,\omega_1]$ is an uncountable compact set and it does not contain any subsets of the form $[\alpha,\omega_1]$, so your argument doesn't work.
For simplicity suppose that we have only two uncountable compact sets (the argument for the general case is a straightforward generalization), $K_0, K_1$. It suffices to show that for any $\xi<\omega_1$ there is a $\gamma \in K_0\cap K_1$ so that $\gamma>\xi$. Fix $\xi<\omega_1$, since $K_0$ is uncountable there is an $\alpha_0\in K_0$ with $\alpha_0>\xi$, by the same argument applied to $K_1$ there is a $\beta_0>\alpha_0\ \beta_0\in K_1$. Using the same argument we can construct recursively sequences $\{\alpha_n\}_{n\in\mathbb N}\subseteq K_0, $ and $\{\beta_n\}_{n\in\mathbb N}\subseteq K_1, $ such that $$\xi<\alpha_0<\beta_0<...<\alpha_n<\beta_n...$$ Observe that $\sup \alpha_n=\sup \beta_n$, take $\gamma=\sup \alpha_n$, it follows that $\gamma\in K_0\cap K_1$ and $\gamma>\xi$, which finishes the proof.