Statement of lemma 1: Let $\delta$ be a proximity relation on the Tychonoff space $X$ corresponding to the compactification $Y$. The compactification $Y$ of the space $X$ is perfect w.r.t to the open set $U \subset X$ iff, for every set $A \subset U$, $A \bar{\delta} \text{Fr}_{X}(U) \implies A \bar{\delta}(X \setminus U).$
Some background info:
We say that the compactification $Y$ of $X$ is perfect w.r.t an open set $U \subset X$ iff the relation $\text{Fr}_{Y}(O \langle U \rangle) = \text{Cl}_{Y}(\text{Fr}_{X}(U)) -(*)$ holds, where
$O \langle U \rangle = Y \setminus \text{Cl}_{Y}(X \setminus U) \\ \text{Fr}_{Y}(O \langle U \rangle) = \text{Cl}_{Y}(O \langle U \rangle) \cap \text{Cl}_{Y}(Y \setminus O \langle U \rangle) $
and $\text{Cl}_{Y}(V)$ is the usual closure of a set $V$ in $Y$. The forward direction of the proof is okay. For the converse, we must show that given any open set $U \subset X$, the relation ($*$) holds.
$\textbf{NB}:$ We will require the following lemma and result from Sklyarenko (both of which I can prove.).
$\textbf{Lemma2:}$ Let $X$ be a Tychonoff space with a compactification $Y$. Let $\delta$ be the proximity relation on $X$ corresponding to $Y$. If $V'$ and $V''$ are two open subsets of $X$, with $V' \bar{\delta} V''$, then $O \langle V' \cup V' \rangle = O \langle V' \rangle \cup O \langle V'' \rangle$ (Note that the proximity relation defined here is given by $V' \bar{\delta} V'' \iff \text{Cl}_{Y}(V') \cap \text{Cl}_{Y}(V'') = \emptyset$).
$\textbf{Result 3}:$ The set $O \langle U \rangle$ is the largest open set in $Y$ such that $O \langle U \rangle \cap X = U.$
Now returning to lemma 1, I can show that the inclusion $\text{Cl}_{Y}(\text{Fr}_{X}(U)) \subset \text{Fr}_{Y}(O \langle U \rangle)$ always holds for any compactification $Y$ of $X$. It is the reverse inclusion I am having difficulty with.
I will layout the proof as given by Sklyarenko and show where I am stuck:
We must show that $\text{Fr}_{Y}(O \langle U \rangle) \subset \text{Cl}_{Y}(\text{Fr}_{X}(U))$. Let us take any $t \in Y$ such that $t \notin \text{Cl}_{Y}(\text{Fr}_{X}(U))$. Then we can find an open set $G$ in $Y$ such that $t \in G$ and
$\text{Cl}_{Y}(G) \cap \text{Cl}_{Y}(\text{Fr}_{X}(U)) = \emptyset - (**)$.
Let $V' = G \cap U$ and $V'' = G \cap (X \setminus \text{Cl}_{X}(U))$.
Then $V'$ and $V''$ are open in $X$. We also have that since $G \cap \text{Fr}_{X}(U) = \emptyset$, we have that
$G \cap X = V' \cup V''$
[this is true since $U$ is open in $X$, then $\text{Fr}_{X}(U) = \text{Cl}_{X}(U) \setminus U$ and we can write $X = U \cup (X \setminus \text{Cl}_{X}(U)) \cup (\text{Cl}_{X}(U) \setminus U)$]. However, by the Result 3, we must have that $G \subset O \langle V' \cup V'' \rangle.$
So we have $t \in G \implies t \in O \langle V' \cup V'' \rangle$. Now, since we have that $\text{Cl}_{Y}(G) \cap \text{Cl}_{Y}(\text{Fr}_{X}(U)) = \emptyset $, it follows that $\text{Cl}_{Y}(V') \cap \text{Cl}_{Y}(\text{Fr}_{X}(U)) = \emptyset$. Hence we have $V' \bar{\delta} \text{Fr}_{X}(U)$ and by our assumption it follows that $V' \bar{\delta} (X \setminus U)$, thus $V' \bar{\delta} V''.$
To see that $V' \bar{\delta} V''$, suppose there exists an $x \in \text{Cl}_{Y}(V') \cap \text{Cl}_{Y}(V'')$, then $x \in \text{Cl}_{Y}(V')$ and $x \in \text{Cl}_{Y}(V'')$. So $x \in \text{Cl}_{Y}(G \cap (X \setminus \text{Cl}_{X}(U))) \subset \text{Cl}_{Y}(G) \cap \text{Cl}_{Y}(X \setminus (\text{Cl}_{X}(U)) -(***)$.
Claim: $x \in \text{Cl}_{Y}(X \setminus U)$. To see this, suppose $x \notin \text{Cl}_{Y}(X \setminus U)$, then there exists an open set $H$ in $Y$ such that $x \in H$ and $H \cap (X \setminus U) = \emptyset$. However, this implies that $H \cap (X \setminus \text{Cl}_{X}(U)) = \emptyset$ (since $X \setminus \text{Cl}_{X}(U) \subset X \setminus U$). Thus we have $x \notin \text{Cl}_{Y}(X \setminus \text{Cl}_{X}(U))$, a contradiction to $(***)$. Hence $x \in \text{Cl}_{Y}(X \setminus U)$ and so $x \in \text{Cl}_{Y}(V') \cap \text{Cl}_{Y}(X \setminus U)$, a contradiction to the fact that $V' \bar{\delta} (X \setminus U)$. Hence $V' \bar{\delta} V''$.
Now, by Lemma 2, we have that $t \in O \langle V' \cup V' \rangle \implies t \in O \langle V' \rangle \cup O \langle V'' \rangle$. The rest of the proof from here is fine as well, so I wont make this any post any longer (sorry).
$\text{TL;DR}$: Here is my question. I am for the most part okay with the proof. My concern is how do we attain the property labelled as $(**)$, as this result plays a crucial role in the proof? Please note that I have filled in a lot of subtleties in the proof, so please feel free to point out any errors in the proof, thank you.