Countable set with singleton is closed is not a pathwise connected

Let $X$ be a countable set with topology $T$ such that the singleton is closed. Recall the following well-known fact. I think it was done by Sierpi'nski.

Fact. $[0,1]$ can not be written as the union of a pairwise disjoint family $\mathcal F$ of nonempty closed sets with $1<|\mathcal F|\le\omega$.

We can conclude that there is no non-constant continuous function $f\colon [0,1]\to X$ since, we have $$[0,1]=\bigcup_{x\in X} f^{-1}\{x\} $$ and $\mathcal F:=\{f^{-1}\{x\}\colon x\in X\}\setminus \emptyset $ is the family of pairwsie disjoint closed sets and $|\mathcal F|\leq\omega,$ which is impossible by the fact above. As consequence, $(X, T)$ is not a pathwise connected.

Q1. Is the above argument correct?

Q2. Is there another way to show that $(X, T)$ is not pathwise connected?

Solution 1:

Yes, the use of Sierpiński's theorem is correct. And IMHO it's essentially the only way to show this general fact in an easy way. Of course if we know the countable space is moreover metrisable or normal there are easier arguments as then $X$ cannot even be connected, e.g. But only knowing $T_1$ for $X$ this approach is straightforward and direct.

BTW the Sierpiński theorem holds for any compact connected Hausdorff space (a continuum), not just $[0,1]$.