$\vert\{\alpha\in I:F_\alpha\cap V\neq\emptyset\}\vert<\aleph_0\implies \bigcup\limits_{\alpha\in I} F_\alpha$ is closed. Let $X$ be a topological space and $\{F_\alpha:\alpha\in I\}$ a family of closed subsets of $X.$ If for all $x\in X,$ there exists $V\in\mathcal V(X)$ such that $\vert\{\alpha\in I:F_\alpha\cap V\neq\emptyset\}\vert<\aleph_0$, then $\displaystyle\bigcup_{\alpha\in I} F_\alpha$ is closed.
I have no much idea on how to prove it.
I've checked others questions and found only one similar 
Natural conditions implying the union of closed sets is closed and I think the conversely of Alex Ravinsky's answer has something to do with my statement:

Conversely, if a point $x$ has a neighborhood $O_x$ intersecting only finitely many members of the family $\mathcal F$ then $O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$ is an open neighborhood of the point $x$ disjoint from $F$.

However I could not see how the arguments induce to the desired result $\displaystyle\bigcup_{\alpha\in I} F_\alpha$ is closed.
Can someone please help me with this difficult exercise?
Thanks in advance.
 A: 
Conversely, if a point $x$ has a neighborhood $O_x$ intersecting only finitely many members of the family $\mathcal F$ then $O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$ is an open neighborhood of the point $x$ disjoint from $F$.

This shows that every point $x\in X\setminus F$ has an open neighborhood $U_x\subseteq X\setminus F$. (More specifically, $U_x=O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$.)
Consequently, $X\setminus F$ is open, since we get $X\setminus F=\bigcup\limits_{x\in X} U_x$, i.e., it is a union of open sets.
A: Note it suffices to prove that $\overline{\bigcup_{\alpha}F_{\alpha}} = \bigcup_{\alpha}\overline{F}_{\alpha}$, as each $F_{\alpha}$ is closed.
The inclusion $\bigcup_{\alpha}\overline{F}_{\alpha} \subset \overline{\bigcup_{\alpha}F_{\alpha}}$ is clear.
Conversely, if $x \in \overline{\bigcup_{\alpha}F_{\alpha}}$ and not in $\bigcup_{\alpha}\overline{F}_{\alpha}$, take $V$ to be an open neighbourhood of $x$ that touches only finitely many $F_{\alpha}$, say $F_1,...,F_n$.
Note $x \notin F_{\alpha}$ for any $\alpha$ so $x \in \bigcap_{i=1}^n X - F_i$ and therefore in $V \cap \bigcap_{i=1}^n X - F_i$ which is open and touches none of the $F_{\alpha}$ - contradiction.
A: As the empty set is in V(X), there is a counter example.  
Let F$_j$ = [-j,j] for all j in (0,1) and V be the empty set.
Though V intersects the F's finite (zero) many times,
$\cup${ F$_j$ : j in (0,1) } = (-1,1) is not closed.  
That "for all x in X" is superfluous as x is not ever used.
Perhaps you were asking about locally finite collections.
Collections with for all x, exists open set V with x in V and V intersecting the F's finitely many times.
Using V(X) instead of the topology is useless complexity.
