The union of a locally finite family of functionally closed- RYSZARD ENGELKING In the book RYSZARD ENGELKING General Topology he states

Show that the union of a locally finite family of functionally closed sets is not necessarily functionally closed. Note that in a perfectly normal space the union of a locally finite family of functionally closed sets is functionally closed.

A topological space $X$ is called a perfectly normal space if $X$ is a normal space and every closed subset of $X$ is $G_{\delta}$-set.
A subset $A$ of a topological space $X$ is called functionally closed if $A=f^{-1}(0)$ for some continous function $f:X \to [0,1]$
Every closed subset of $X$ is $G_{\delta}$-set means that it countable intersections of open sets
On the other hand,
observation 1: I have proven that a subset $A$ of a normal space is closed $G_{\delta}$ if and only if there exists a continuous function $f:X \to [0,1]$ such that $A=f^{-1}(0)$. i.e, in a normal space functionally closed sets coincide with closed $G_{\delta}$.
A family $\{A_s\}_{s \in S}$ of subsets of a topological space $X$ is locally finite if for every point $x \in X$ there exists a neighbourhood $U$ such that the set $\{s \in S : U \cap A_s \neq \emptyset \}$ is finite.
Any ideas:
Suposse that $X$ is perfectly normal espace.
Let $A=\{A_s\}_{s \in S}$ a locally finite family of functionally closed sets, for observation 1, the $A_s$ are $G_{\delta}$-sets.  We have to find a continuous function $f:X \to [0,1]$ such that $\cup_{s \in S} A_s= f^{-1}(0)$. For each $A_s$ there is a continuous function $f_s: X \to [0,1]$ such that $f_s^{-1}(0)=A_s$.I don't know how to use my hypotheses to build my continuous function f? some help?
On the other hand, I can't think of a counterexample for the first part.
 A: As you discovered, the set $\Bbb Q$ in the space that I constructed in my answer to this question is the union of a closed, discrete (hence locally finite) family of functionally closed sets but is not itself functionally closed.
For the rest of the question, the union of a locally finite family of closed sets is closed, so if the space is perfectly normal, the union is a $G_\delta$-set and therefore functionally closed.
A: Let $N=\Bbb R\times [0,\infty)$ with the topology of the Niemitzky plane (a.k.a the  Niemitzky-Moore plane). For brevity let $E=\Bbb R\times [0,\infty).$ Then $E$ is a closed discrete subspace of $N$ so every $A\subset E$ is closed in $N.$ We show that every $A\subset E$ is the intersection of a functionally-closed family in $N,$ and that some subsets of $E$ are not functionally-closed in $N.$
For $(r,0)=p\in E$ and $s\in [0,\infty$ let $B(p,s)=\{p\}\cup \{(x,y)\in N:(x-r)^2+(y-s)^2<s^2\}.$ Then $B(p,s)$ is open, and $\overline {B(p,s)}= B(p,s)\cup \{(x,y)\in N:(x-r)^2+(y-s)^2=s^2\}.$
For $p\in E$ and $p\ne q\in B(p,1)$ there is a unique circle $C$ containing $q$ and tangent to $E$ at $p.$ Let $t$ be the radius of $C$  and let $f_p(q)=1-t.$ For $ q\in N$ \ $B(p,1)$ let $f_p(q)=0.$ Let $f_p(p)=1.$ We can confirm that $f_p:N\to [0,1]$ is continuous by observing that if $0<z<1$ then $f_p^{-1}(z,1]=B(p,1-z)$ is open, and $f_p^{-1}[0,z)=N$ \ $\overline {B(p,1-z)}$ is open.
The "point" is that $f_p^{-1}(0,1]$ is functionlly open, and its intersection with $E$ is $\{p\}.$
Now  $N$ \ $E$ is functionally open. So for any $A\subset E,$ the set $F_A=\{N \backslash E\}\cup \{f_p^{-1}(0,1]:p\in E\backslash A\}$ is a functionally open family. If $\cup F_A$ is functionally open then $A=N$ \ $\cup F_A$ is functionally closed.
Let $C(N, \Bbb R)$ be the set of continuous $f:N\to \Bbb R.$ Since $\Bbb R$ is Hausdorff and $N$ is separable, we have $|C(N,\Bbb R)|\le |\Bbb R|^{\aleph_0}=2^{\aleph_0}.$ But if every $A\subset E$ were functionally closed in $N,$ we would have $|C(N,\Bbb R)|\ge |\{A:A\subset E\}|=2^{2^{\aleph_0}}.$
