Compactifications with remainder $[0,\omega_1]$ and convergent sequences

Question

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 has negative answer.

Answer 1

Here's an example, suggested by Alan Dow. Take a 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 $\omega_1$.