A compact, Hausdorff space $X$ is sequential if each for each $A\subset X$ and $x\in \overline{A}$, there exists a countable set $A_0\subset A$ such that $x\in \overline{A}_0$.
I am asked to show that each sequential, compact Hausdorff space without isolated points of weight $\aleph_1$ contains a closed zero-dimensional subspace $X_0$ of weight $\aleph_1$.
Here's my attempt. let $\{U_\alpha\colon \alpha<\omega_1\}$ be a basis for $X$. Inductively choose a one-to-one trasnfinite sequence $x_\alpha$ such that $x_\alpha\in U_\alpha\setminus U_\beta$ for all $\beta<\alpha$. Is $X_0 = \overline{\{x_\alpha\colon \alpha<\omega_1\}}$ the right choice?