Countable union covers a second countable, compact Hausdorff space

Question

I think that the following claim is wrong, but I could not come up with a counter example:

Let $X$ be a compact, Hausdorff, second countable space.

Assume that we have the following process: Let $U_1\subseteq X$ be a given open dense subset. Let $U_2\subseteq X\backslash U_1$ be a given relatively open, dense subset. Let $U_3\subseteq X\backslash (U_1\cup U_2)$ be a given relatively open, dense subset. Coninue with this process.

Does it follow that $X=\bigcup_{i=1}^{\infty} U_i$?

Answer 1

One possibility is a scattered space of rank $> ω$.

Another one are sets $(∏_{k < n} \{1\}) × (S^1 \setminus \{1\}) × (∏_{k > n} S^1)$ for $n ∈ ω$ in $(S^1)^ω$.

Answer 2

No. A counter-example: Let $X$ be the ordinal $\omega^{\omega}+1$ with the $\epsilon$-order topology. Then $X$ is compact Hausdorff. And $X$ is homeomorphic to a sub-space of $\Bbb R.$ So $X$ is second-countable. Let $U_1=\{0\}\cup \{e+1:e\in X\}$ and $U_{n+1}=\{\omega^n x: 0\in x\in \omega^{\omega}\}.$ The ordinal $\omega^{\omega}$ is a member of $X$ which does not belong to $U_n$ for any $n\in \Bbb N.$

Remarks. $U_1$ is the set of isolated points of $X$ and $U_{n+1}$ is the set of isolated points of the space $X\setminus (\cup_{j=1}^nU_j).$