Proof of lemma 2 of "some questions in the theory of bicompactifications" I am having trouble proving the first claim made in the proof of the following lemma:
Lemma 2: Let $X$ be a completely regular space with the bicompactification $Y$. Let $V'$ and $V''$ be two open subsets of $X$ with $\operatorname{Cl}_{Y} V' \cap \operatorname{Cl}_{Y} V'' = \emptyset$. Then
\begin{align*}
Y \setminus \operatorname{Cl}_{Y} (X \setminus (V' \cup V'')) = [Y \setminus \operatorname{Cl}_{Y} (X \setminus V')] \cup [Y \setminus \operatorname{Cl}_{Y} (X \setminus V'')]
\end{align*}
Proof: Let $V = V' \cup V''$. 
Claim: $V$ is dense in $Y \setminus \operatorname{Cl}_{Y} (X \setminus V)$.
We must prove that $\operatorname{Cl}_{Y} V = Y \setminus \operatorname{Cl}_{Y} (X \setminus V) $. Now, I have been able to verify the inclusion
\begin{align*}
Y \setminus \operatorname{Cl}_{Y} (X \setminus V) \subset \operatorname{Cl}_{Y} V
\end{align*}
for if $x \in Y \setminus \operatorname{Cl}_{Y} (X \setminus V) \Rightarrow x \notin  \operatorname{Cl}_{Y} (X \setminus V) \supset X \setminus V \Rightarrow x \notin X \setminus V \Rightarrow x \in V \subset \operatorname{Cl}_{Y} V.$ For the reverse inclusion, I am having difficulty. Let $x \in \operatorname{Cl}_{Y}V$ and suppose that $x \in \operatorname{Cl}_{Y} (X \setminus V).$ I am trying to find a contradiction to prove that $x \notin \operatorname{Cl}_{Y} (X \setminus V)$ and I have a hunch that what we were given ($\operatorname{Cl}_{Y} V' \cap \operatorname{Cl}_{Y} V'' = \emptyset$) might be the way or some set-theoretic manipulation ? 
P.S:  I have not used the fact that $X$ is completely regular as it is assumed all spaces are completely regular (so that we can have compactifications).
Any assistance with the reverse inclusion is appreciated.
 A: This technical question has a simple technical answer
$$Y \setminus \operatorname{Cl}_{Y} (X \setminus (V' \cup V'')) = [Y \setminus \operatorname{Cl}_{Y} (X \setminus V')] \cup [Y \setminus \operatorname{Cl}_{Y} (X \setminus V'')]$$
$$ \operatorname{Cl}_{Y} (X \setminus (V' \cup V'')) = \operatorname{Cl}_{Y} (X \setminus V') \cap \operatorname{Cl}_{Y} (X \setminus V'').$$
The inclusion $\subset$ is obvious. To show the converse inclusion, assume that $y\not\in \operatorname{Cl}_{Y} (X \setminus (V' \cup V''))$. Then there exists an open in $Y$ set $U\ni y$ such that $U\cap (X \setminus (V' \cup V''))=\varnothing$. That is, $U\cap X\subset V' \cup V''$. Since $\operatorname{Cl}_{Y} V' \cap \operatorname{Cl}_{Y} V'' = \varnothing$, in order to show that $y\not\in \operatorname{Cl}_{Y} (X \setminus V') \cap \operatorname{Cl}_{Y} (X \setminus V'')$
without loss of generality we may assume that $y\not\in \operatorname{Cl}_{Y} V''$. There exists an open in $Y$ neighborhood $U’$ of the point $y$ such that $U’\cap V’’=\varnothing$.
Then $U\cap U'\cap X\subset V'$, that is $y\not\in \operatorname{Cl}_{Y} (X \setminus V')$.
