A question about the closedness of $C$-embedded subsets

Question

In the answer of this post it is proven that in every first countable Tychonoff space the $C^{*}$-embedded subsets are closed. Since countable pseudocharacter (i.e., points are $G_\delta$ sets) is a natural weakening of first countability, the corresponding question surges immediately: is it true that in every Tychonoff space with countable pseudocharacter the $C^*$-embedded subsets are closed?

Now, it is not difficult to prove that the Single Ultrafilter Topology is a Tychonoff space with countable pseudocharacter that admits a non-closed $C^*$-embedded subset. However, it can be checked that the natural subset of the preceding space that checks the properties in question is not $C$-embedded in the space. For this reason, a natural question arises again: is it true that in Tychonoff spaces with countable pseudocharacter it is satisfied that the $C$-embedded subsets are closed?

Regarding this last question, it appears as exercise 1K(2) in the book "Extensions and Absolutes of Hausdorff spaces" by Porter and Woods; therefore, it seems that the answer to this last question is affirmative. However, I have not been able to prove it. Any kind of help to find out the truth would be greatly appreciated!

Answer 1

Suppose $X$ is Tychonoff with countable pseudocharacter and $A\subset X$ is not closed. Let $x\in \overline{A}\setminus A$. Since the singleton $\{x\}$ is $G_\delta$, it is a zero set, i.e. there exists $f:X\to\mathbb{R}$ which vanishes at $x$ and nowhere else. The function $1/f$ is then well-defined and continuous on $A$ but does not extend continuously to $X$, since it must approach $\pm\infty$ as you approach $x$ in $A$.