Help understanding a proof about $C^*$-embedded subsets In this link the author of the post published a short proof showing that every $C^*$-embedded subset in a first countable Tychonoff space is necessarily closed. There comes a point in the proof where the author claims that a certain function $\hat{g}f:S\to[0,1]$ does not admit a continuous extension to the space $S\cup\{x_0\} $. Although there is the possibility that this fact is a triviality, I have not been able to verify that this is indeed the case. Could any of you please help me understand why that little step is true?
 A: Recall that a necessary and sufficent condition for a subset $S\subseteq X$ to be $C^*$-embedded is that any two disjoint zero-sets of $S$ are completely separated in $X$.

Proposition: Asume that $X$ is a first-countable Tychonoff space and $S\subseteq X$ its $C^*$-embedded subspace. Then $S$ is closed in $X$.

Proof: In case $S$ is not closed there is a sequence of distinct points $\{x_n\}_\mathbb{N}\subseteq S$ such that $x_n\rightarrow p\in\overline S\setminus S$. Put
$$A=\{x_{2n}\mid n\in\mathbb{N}\},\qquad B=\{x_{2n+1}\mid n\in\mathbb{N}\}.$$
Then $A\cup\{p\}$ and $B\cup\{p\}$ are easily seen to be compact $G_\delta$-sets in $X$. Since $X$ is Tychonoff, $A\cup\{p\}$ and $B\cup\{p\}$ are zero-sets in $X$. Consequently $A,B$ are disjoint zero-sets in $S$. Evidently they are not completely separated in $X$, and this contradicts the assumption that $S$ is $C^*$-embedded. $\quad\blacksquare$
As for the original question, we have the following

Theorem (Taimanov): Let $Y$ be a compact $T_2$ space and $X$ any space. Suppose that $D\subseteq X$ is a dense subspace and $f:D\rightarrow Y$ a continuous function. Then $f$ extends continuously over $X$ if and only if $cl_X(f^{-1}(A))\cap cl_X(f^{-1}(B))=\emptyset$ for any two disjoint closed subsets $A,B\subseteq Y$. $\quad\blacksquare$

As per the linked post we have that $S$ is dense in $X=S\cup\{p\}$, but with $h=\hat g\circ f:X\rightarrow [0,1]$ we get $cl_X(h^{-1}(\{0\}))\cap cl_X(h^{-1}(\{1\}))=\{p\}\neq\emptyset$.
