Countable union of closed subsets of a pseudoradial and sequentially closed topological space is closed

Question

A topological space $X$ is called pseudoradial if, for every non-closed subset $A$ of $X$, there exists $p\in \overline{A}\setminus A$ and a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ that converges to $p$. A topological space $X$ is called sequentially discrete if every convergent sequence in the space is eventually constant.

Question: If a topological space $X$ is at the same time pseudoradial and sequentially discrete, must a countable union of closed subsets of $X$ always be closed?

Easy observation. (From @PatrickR) A countable union of singletons, namely a countable subset of $X$, is always closed. Suppose that $A$ is a countable non-closed subset of $X$, then by definition of the pseudoradial property, there exists $p\in \overline{A}\setminus A$ and a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ in $A$ convergeing to $p$. Since $A$ is countable, this answer shows that there exists in fact an ordinary sequence of points from $A$ converging to $p$, which contradicts sequential discreteness of $X$.

Thank you for any help in advance.

Answer 1

Yes:

First observe that
($*$) if $(x_\alpha)_{\alpha < \lambda}$ converges to $p$, and $T$ is cofinal in $\lambda$, then also $(x_\alpha)_{\alpha \in T}$ converges to $p$.

Now let $(A_n)_{n \in \mathbb N}$ be a family of closed sets of $X$. Assume $A = \bigcup _{n \in \mathbb N} A_n$ is not closed. Then there exists $p\in \overline{A}\setminus A$ and a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ that converges to $p$. By ($*$) we may assume that $\lambda$ is regular. Since $X$ is sequentially discrete, $\lambda$ is uncountable.

For $n \in \mathbb N$, let $T_n = \{\alpha < \lambda: x_\alpha \in A_n\}$. Then there exists $n \in \mathbb N$, such that $T_n$ is cofinal in $\lambda$. Hence, again by ($*$), $(x_\alpha)_{\alpha \in T_n}$ converges to p, hence $p \in A_n$. Contradiction!