I am stuck trying to prove some result from topology. It is fairly complicated to state it here, but I realized that I could prove it, if I manage to show the following Lemma.
Suppose $X$ is a normal space (if necessary, we can even assume perfectly normal). Suppose $A$ is a closed subspace of $X$ and $B$ is a closed subspace of $A$, so $$ B \subseteq A \subseteq X. $$ Furthermore, suppose there is some other space $C \subseteq X$, for which we know that $C \cap A = \emptyset$ and $\overline{C} \cap A \subseteq B$. Can we find an open neighborhood $W$ of $C$ for which $\overline{W} \cap A \subseteq B$?
If we can, how can we prove its existence? If not, could you please provide a counterexample?
This is not true if $X$ is merely normal (even under the Hausdorff condition). A counterexample is the Tychonoff Plank $X=[0,\omega_1]\times [0,\omega]$, with the product of the order topologies. Let $A=[0,\omega_1]\times \{\omega\} $, $B=\{(\omega_1,\omega)\}$, and $C=\{\omega_1\}\times [0,\omega)$.
Then every neighborhood $W$ of $C$ has $W\supseteq [\alpha,\omega_1]\times [0,\omega)$ for some countable ordinal $\alpha<\omega_1$, so $\overline{W}\cap A\supseteq [\alpha,\omega_1]\times \{\omega\}$.
On the other hand, if $X$ is completely normal (i.e., "hereditarily normal"), then you can indeed find such a $W$. By normality of $X\backslash B$, since $\overline{C}\backslash B$ and $A\backslash B$ are (relatively) closed and disjoint, there is a (relatively) open set $W\subset X\backslash B$ with $C\subseteq \overline{C}\backslash B\subseteq W$ and $(\overline{W}\backslash B)\cap (A\backslash B)=\emptyset$, whereby $\overline{W}\cap A\subseteq B$. Since $B$ is closed, $W$ is in fact open in $X$ as well, and is thus the desired set.
Note that complete normality is weaker than perfect normality.
Remark.
In fact, a converse is true: your lemma is equivalent the property that $X$ is completely normal. To see this, suppose your lemma is true for a space $X$, and let $Y\subseteq X$. Then let $C,A_1\subset Y$ be (relatively) closed and disjoint. Let $A=\overline{A_1}$, where the closure is taken with respect to $X$, and let $B=A\cap \overline{C}$. Then $B\subseteq A$ are closed, and $\overline{C}\cap A\subseteq B$ and $C\cap A=\emptyset$, so $A$, $B$, and $C$ satisfy the assumptions of the lemma.
Then the resulting $W$ satisfies $W\cap Y\supseteq C$ and $\overline{W}\cap Y \cap A_1=\emptyset$ (since $\overline{W}\cap A_1\subseteq \overline{W}\cap A\subseteq B$, and $B\cap Y=\emptyset$), and since $A_1$ and $C$ were arbitrary closed subsets of $Y$, $Y$ is normal. Since this holds for all subsets $Y\subseteq X$, $X$ is completely normal.