Let $(X, \tau)$ be a topological space. By $T_3$ I will mean that $\forall x \in X$ and $A $ closed subset of $X$ that does not contain $x$, there exists disjoint open sets $U, V$ such that $x \in U$, $A \subset V$. The following exercise appears in Dugundji (p. 157): If each point $x$ of $X$ has a closed neighborhood $C$ such that $C$ equipped with the subspace topology is $T_3$ then $X$ is $T_3$. I think I have proved the result, but it bugs me that I did not use the fact that $C$ is closed. The proof goes as follows. Let $x$ and $C$ be as in the hypothesis and let $U \in \tau$ such that $x \in U$. Because $C$ is $T_3$, there exists $V \in \tau$ such that
$$ x \in C \cap V \subset \mathcal{C}_{C} (C \cap V) \subset C \cap U.$$
Where $C_{A}$ denotes the closure relative to a subset $A$ of $X$. Because $C$ is an neighborhood of $x$, we get $x \in \mathring{C}$. Now, if $y \in \mathcal{C}_{X} (V \cap \mathring{C})$, for every $K \in \tau$ such that $y \in K$, we get that
$$ K \cap V \cap \mathring{C} \neq \emptyset$$ which imples $$ (K \cap C) \cap V \cap C \neq \emptyset $$ by this mean, we obtain that $ y \in \mathcal{C}_{C} (V \cap C )$, i.e. $\mathcal{C}_{X} (V \cap \mathring{C}) \subset \mathcal{C}_{C} (V \cap C )$. Finally, $V \cap \mathring{C}$ is an open neighborhood of $x$ and satisfies $$ \mathcal{C}_{X}(V\cap \mathring{C}) \subset\mathcal{C}_{C} (C \cap V) \subset C \cap U \subset U .$$ By a characterization of $T_3$ spaces we get that $X$ is $T_3$. I was wondering if someone could point out where I went wrong (if I did) or show me a proof that uses that $C$ is closed.
I'll present the way I'd do it first. It is highly similar to your approach.
Fix such a nonempty space $X$; fix $x\in X$, $A\subseteq X$ closed and not containing $x$. Pick such a closed neighbourhood $C$ of $x$.
$K:=A\cap C$ is closed in $C$. Since $C$ is a $T_3$ subspace, there is a $C$-open neighbourhood $V$ of $x$ whose $C$-closure is disjoint from $K$. As $C$ is a neighbourhood of $x$ in $X$, we can take $V$ to be open in $X$ too.
Suppose there exists an element $y\in\overline{V}\cap A$. Necessarily, $y$ is not in $C$ (check this carefully). Because $C$ is closed, $U:=X\setminus C$ is an open neighbourhood of $y$. Since $y\in\overline{V}$, $U\cap V$ is nonempty. However, since $V\subseteq C$, this is a contradiction.
Therefore, $\overline{V}\cap A$ is empty. It follows that $X$ is $T_3$.
As for your proof, the mistake is in inferring $y\in\mathcal{C}_C(V\cap C)$. This would be true... if $y\in C$. $C\cap\mathcal{C}_X(V\cap C)\subseteq\mathcal{C}_C(V\cap C)$ is true but if you remove the "$C\cap\cdots$" then it might not be true. However, since $C$ is closed, if $y\in\mathcal{C}_X(V\cap C^{\circ})$ then $y\in\mathcal{C}_X(C)=C$, so all is well.