Is the following statement consistent?


$(\star)$ Let $K$ be a separable compact Hausdorff space containing the space $[0,\omega\_1]$ so that the complement $K\setminus[0,\omega\_1]$ is discrete. Then there exists an open neighborhood $U$ of $[0,\omega\_1)$ in $K$ such that $U$ contains no sequences convergent to $\omega\_1\in[0,\omega\_1]$.


**Remark 1.** The statement $(\star)$ does not hold under $\omega\_1<\mathfrak p$. That is why I am asking only about the consistency of $(\star)$.


**Remark 2.** If $(\star)$ is consistent, then Question 1 [in this MO-post](https://mathoverflow.net/q/351527/61536) has negative answer.



Compactifications with remainder $[0,\omega_1]$ and convergent sequences
Here's an example, suggested by Alan Dow. Take a [Hausdorff Gap](https://en.wikipedia.org/wiki/Hausdorff_gap): a pair of sequences $\langle a\_\alpha:\alpha<\omega\_1\rangle$ and $\langle b\_\alpha:\alpha<\omega\_1\rangle$ of infinite subsets of $\mathbb{N}$ such that $a\_\alpha\subset^\*a\_\beta$ and $b\_\alpha\subset^\*b\_\beta$ whenever $\alpha<\beta$ and $a\_\alpha\cap b\_\alpha=^\*\emptyset$ for all $\alpha$, and with the property that whenever $A$ is such that $a\_\alpha\subseteq^\*A$ for all $\alpha$ there is an $\alpha$ such that $A\cap b\_\alpha$ is infinite.
Topologise $K=\mathbb{N}\cup[0,\omega\_1]$ by the subbase consisting of $\{n\}$ and $K\setminus\{n\}$ for all $n\in\mathbb{N}$, as well of the sets $a\_\alpha\cup[0,\alpha]$ and $(\mathbb{N}\setminus a\_\alpha)\cup(\alpha,\omega\_1]$ for all $\alpha$.
This yields a compactification of $\mathbb{N}$ with $[0,\omega\_1]$ as its remainder in which every set $b\_\alpha$ converges to $\omega\_1$.
If $U$ is open in $K$ and $U\cap[0,\omega\_1]=[0,\omega\_1)$ it then follows that $a\_\alpha\subseteq^\*U\cap\mathbb{N}$ for all $\alpha$. But then for some $\alpha$ the set $U\cap b\_\alpha$ is infinite and it converges to $\omega\_1$. 


