[proofplan] The complement $X \setminus W$ is closed and disjoint from $A$. Normality separates these two closed sets by disjoint open neighbourhoods $U$ and $V$, with $A \subset U$ and $X \setminus W \subset V$. The disjointness forces $U \subset X \setminus V$, and since $X \setminus V$ is closed, the closure $\overline{U}$ is still contained in $X \setminus V$. Finally, $X \setminus V \subset W$, so this [open set](/page/Open%20Set) $U$ is the desired shrinking of $W$ around $A$. [/proofplan] [step:Separate $A$ from the closed complement of $W$] Since $W \in \tau$, the set $F := X \setminus W$ is closed in $(X,\tau)$. Since $A \subset W$, we have \begin{align*} A \cap F = A \cap (X \setminus W) = \varnothing. \end{align*} Thus $A$ and $F$ are disjoint closed subsets of $X$. By normality of $(X,\tau)$, there exist open sets $U,V \in \tau$ such that \begin{align*} A \subset U, \quad F \subset V, \quad U \cap V = \varnothing. \end{align*} [guided] We want an open set $U$ containing $A$ whose closure remains inside $W$. The obstruction to being inside $W$ is precisely the complement of $W$, so define \begin{align*} F := X \setminus W. \end{align*} Because $W$ is open in the topology $\tau$, its complement $F$ is closed in $(X,\tau)$. Because $A \subset W$, no point of $A$ lies in $X \setminus W$, hence \begin{align*} A \cap F = A \cap (X \setminus W) = \varnothing. \end{align*} Now both sets to be separated are closed: $A$ is closed by hypothesis, and $F$ is closed because it is the complement of the open set $W$. The defining separation property of normality therefore applies to the disjoint closed sets $A$ and $F$. Hence there exist open sets $U,V \in \tau$ such that \begin{align*} A \subset U, \quad F \subset V, \quad U \cap V = \varnothing. \end{align*} The set $U$ is the candidate shrinking around $A$, and $V$ is an open buffer around the part of $X$ that lies outside $W$. [/guided] [/step] [step:Use the closed complement of $V$ to control the closure of $U$] Since $U \cap V = \varnothing$, every point of $U$ lies in $X \setminus V$, so \begin{align*} U \subset X \setminus V. \end{align*} Because $V$ is open, $X \setminus V$ is closed. Since $\overline{U}$ is the smallest closed subset of $X$ containing $U$, and $X \setminus V$ is a closed subset of $X$ containing $U$, it follows that \begin{align*} \overline{U} \subset X \setminus V. \end{align*} [/step] [step:Conclude that the closure of $U$ lies inside $W$] From $F \subset V$ and $F = X \setminus W$, we obtain \begin{align*} X \setminus W \subset V. \end{align*} Taking complements reverses containment, so \begin{align*} X \setminus V \subset W. \end{align*} Combining this with the previous step gives \begin{align*} \overline{U} \subset X \setminus V \subset W. \end{align*} Also $U \subset \overline{U}$ by the definition of closure, and the first step gave $A \subset U$. Therefore \begin{align*} A \subset U \subset \overline{U} \subset W. \end{align*} The set $U$ is open because it was chosen by normality as an element of $\tau$. This completes the proof. [/step]