Uspenskij-Tkachenko Theorem 
Theorem: For a Moscow space $X$, every 
  $G_\delta$-dense subset $Y$ of $X$ is C-embedded in $X$.

Proof $^{1}$: Assume that $Y$ is not C-embedded in $X$. Then, as it is easy to see, there
are open subsets $V_1$ and $V_2$ of $Y$ such that their closures in $Y$ are disjoint, while
the intersection of the closures of $V_1$ and $V_2$ in $X$ is not empty. Fix a point $x$ in
$\overline{V_1}\cap \overline{V_2}$, and let $U_i$ be the interior of the closure of $V_i$ in $X$, $i = 1, 2$. Obviously,
$V_i\subset U_i$; therefore, $U_i$ is not empty.
Since $X$ is a Moscow space, we can find $G_\delta$-sets $P_i$ in $X$ such that $x\in P_i\subset \overline{U_i}$,
$i = 1, 2$. Then $P = P_1\cap P_2$ is a $G_\delta$-subset of $X$ and $x\in P$; therefore, $P\cap Y$ is
not empty. Clearly, every point of $P\cap Y$ belongs to the intersection of closures
of the sets $V_1$ and $V_2$ in $Y$ , which is impossible, since this intersection is empty,
by the choice of $V_1$ and $V_2$.


*

*A.V. Arhangel'skii, On a theorem of W.W. Comfort and K.A. Ross, Comment.Math.Univ.Carolin. 40,1 (1999)133–151; dml.cz. 


In Topological Groups and Related Structures , Theorem 6.1.7. p348, he wrote more details for this theorem.

Since $Y$ is not C-embedded in $X$ then $Y$ is not $C^{*}$- embedded in $X$, and Urysohn extension theorem implies that $Y$ contains two completely separated 
subsets $A$ and $B$ whose closures in $X$ intersect. Take open subsets $V_1$ and $V_2$ in $Y$ such that 
$A\subset V_1$, $B\subset V_2$ and the closures of $V_1$ and $V_2$ in $Y$ are disjoint. Clearly, the intersection 
of the closures of $V_1$ and $V_2$ in $X$ is not empty.
How can we take open subset open subsets $V_1$ and $V_2$ of $Y$ such that their closures in $Y$ are disjoint?
Thanks.
 A: $\newcommand{\cl}{\operatorname{cl}}$Suppose that $Y$ is not $C$-embedded in $X$, and let $f:Y\to\Bbb R$ be a continuous function with no continuous extension to $X$. Then there is a $p\in X\setminus Y$ such that $f$ has no continuous extension to $Y\cup\{p\}$. For $n\in\Bbb Z^+$ let $C_n=\cl_Xf^{-1}\big[[-n,n]\big]$; clearly $Y\subseteq\bigcup_{n\in\Bbb Z^+}C_n$. If $p\notin\bigcup_{n\in\Bbb Z^+}C_n$, then $X\setminus\bigcup_{n\in\Bbb Z^+}C_n$ is a $G_\delta$ containing $p$ and disjoint from $Y$, contradicting the fact that $Y$ is $G_\delta$-dense in $X$, so $p\in C_n$ for some $n\in\Bbb Z^+$.
Suppose that $[a,b]$ is a closed interval in $\Bbb R$ with $a<b$ and $p\in\cl_Xf^{-1}\big[[a,b]\big]$. Let $c=\frac12(a+b)$; then $p\in\cl_Xf^{-1}\big[[a,c]\big]$ or $p\in\cl_Xf^{-1}\big[[c,b]\big]$. Thus, repeated bisection starting with $I_0=[-n,n]$ yields a sequence $\langle I_k:k\in\omega\rangle$ of non-degenerate closed intervals in $\Bbb R$ such that $p\in\cl_Xf^{-1}[I_k]$ for each $k\in\omega$, and $\bigcap_{k\in\omega}I_k=\{a\}$ for some $a\in\Bbb R$. Note that if $U$ is any open nbhd of $a$ in $\Bbb R$, then $I_k\subseteq U$ for some $k\in\omega$, and $p\in\cl_Xf^{-1}[U]$.
Define $g:Y\cup\{p\}\to\Bbb R$ by letting $g\upharpoonright Y=f$ and $g(p)=a$; $g$ is discontinuous at $p$, so there is an $\epsilon>0$ such that for each open nbhd $U$ of $p$ in $X$ there is a $y\in U\cap Y$ such that $|g(y)-a|>\epsilon$. Let $V$ and $W$ be open sets in $X$ such that $$V\cap Y=\{y\in Y:|f(y)-a|>\epsilon\}$$ and $$W\cap Y=\left\{y\in Y:|f(y)-a|<\frac{\epsilon}2\right\}\;.$$ Then $p\in\cl_XV\cap\cl_XW$, and $Y\cap\cl_XV\cap\cl_XW=\varnothing$.
