Lemma 1: Let $\delta$ be a proximity relation on the space $X$ corresponding to the bicompactification $Y$. The bicompactification $Y$ of the space $X$ is perfect with respect to the open set $U$ of $X$ if and only if for every set $A \subset U$, $A \bar{\delta} (X \setminus U)$ follows from $A \bar{\delta} \operatorname{Fr}_{X} U$.
I have started the forward direction of the proof and will provide my proof below. I need some clarification regarding a subset inclusion, which will be denoted by ($\star$).
Proof:
Remark: We have that: \begin{align*} & A \bar{\delta} (X \setminus U) \equiv \operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (X \setminus U) = \emptyset \\ & A \bar{\delta} \operatorname{Fr}_{X} U \equiv \operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U) = \emptyset \end{align*} where $\operatorname{Cl}_{Y}A$ and $\operatorname{Fr}_{X} U = \operatorname{Cl}_{X} U \cap \operatorname{Cl}_{X} (X \setminus U)$ denote the closure and frontier (boundary) resp.
If the relation $\operatorname{Fr}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U)) = \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U)$ holds for the open set $U$ of $X$, then we say that the bicompactification $Y$ is perfect with respect to the set $U$.
Let the compactification $Y$ be perfect with respect to the open set $U$ and let $A \subset U$ and $A \bar{\delta} \operatorname{Fr}_{X} U$, i.e \begin{align*} \operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U) = \emptyset \end{align*} We must show that $A \bar{\delta} (X \setminus U)$. Suppose to the contrary that $A \delta (X \setminus U)$, i.e \begin{align*} \operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (X \setminus U) \neq \emptyset \end{align*} Now, pick any $x \in \operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (X \setminus U)$. Then $x \in \operatorname{Cl}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U))$, since \begin{align*} A \subset U \subset Y \setminus \operatorname{Cl}_{Y} (X \setminus U) \quad \textbf{($\star$)} \end{align*} We also have that $x \notin Y \setminus \operatorname{Cl}_{Y} (X \setminus U)$, since $x \in \operatorname{Cl}_{Y} (X \setminus U)$. Hence, we have that $x \in \operatorname{Fr}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U))$, since \begin{align*} \operatorname{Fr}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U)) = \operatorname{Cl}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U)) \cap \operatorname{Cl}_{Y} (X \setminus U) \end{align*} Moreover, we have that $x \notin \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U)$, since $x \in \operatorname{Cl}_{Y} A$ (by choice of $x$) and $\operatorname{Cl}_{Y} A \cap \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U) = \emptyset$.
Thus we have that $x \notin \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U)$ and $x \in \operatorname{Fr}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U))$. Hence $\operatorname{Fr}_{Y} (Y \setminus \operatorname{Cl}_{Y} (X \setminus U)) \neq \operatorname{Cl}_{Y} (\operatorname{Fr}_{X} U)$, a contradiction to the fact that $Y$ is a perfect compactification with respect to $U$.
Question: I am having difficulty showing that $(\star)$ is true. We know that $U \subset X \subset Y$. I tried to use the result below:
Proposition: Let $X$ be a top. space and $A \subset X$. Then, $y \in \operatorname{Cl}_{X} A$ if and only if for every open $G$ in $X$, $y \in G$, then $G \cap A \neq \emptyset$.
Can we say that since $X$ is a dense subspace of $Y$ ($Y$ is a compactification of $X$) and so any open set $U$ of $X$ is of the form $U = G \cap X$, where $G$ is open in $Y$. Then, if $y \in U \Rightarrow y \in Y$. Suppose $y \in \operatorname{Cl}_{Y} (X \setminus U)$. Now, $y \in G \cap X \subset G \Rightarrow y \in G \Rightarrow G \cap (X \setminus U) \neq \emptyset \Rightarrow G \cap (X \setminus (G \cap X)) \neq \emptyset \Rightarrow G \cap (X \setminus G) \neq \emptyset$, a contradiction. Hence $y \in Y \setminus \operatorname{Cl}_{Y} (X \setminus U)$