Consider the following statement:
A subset $V$ of a topological space $X$ is a neighborhood of the set $S$ if and only if $V$ is a neighborhood for each point in $S$
Here is a proof:
$\color{green}{\large\Rightarrow}$
Let $X$ be a topological space and $V,S⊆X$, since $V$ is a neighborhood for $S$, by definition of neighborhood of a set there exist an open subset $U$ of $X$such that $S⊆U⊆V$, now choose $x∈S$, using the definition of subset implies $x∈U⊆V$, based on the definition of neighborhood of a point implies $V$ is a neighborhood for every $x∈S$
for the other direction:
$\color{green}{\large\Leftarrow}$
Assume for every $x$ is $S$ , $V$ is neighborhood for $x$, using the definition of neighborhood of a point implies there exist an open subset $U_x$ of $X$ such that for every $x∈S$ we have $x∈U_x⊆V$, hence $S = \bigcup_{x \in S} x⊆\bigcup_{x \in S} U_x⊆V$ , since $U_x$ is open in $X$ for every $ x∈S$, hence their union is also open in $X$, using the definition of neighborhood of a set we conclude that $V$ is a neighborhood of the set $S$.
Here the characterization I've used for $U$ is :
$$ \color{blue}{\bigcup_{x \in S} U_x⊆U}\tag{*}$$ I don't know if this characterization is right or not,and I think there is no require to say $$ \color{blue}{\bigcup_{x \in S} U_x=U}$$ does always hold, that's why I use the more general case which I've mentioned in (*).
My question is that can someone check the proof and the definition I've used for $U$?
I do not see your problem. You have shown that $S \subset \bigcup_{s \in S} U_x \subset V$ and that the union is open. Now define $U = \bigcup_{s \in S} U_x$. This is an open neighborhood of $S$ as desired.
A perhaps simpler approch is this. Let $U$ denote the interior of $V$. Then $x \in U$ for all $x \in S$ because for each such $x$ there exists an open $U_x$ with $x \in U_x \subset V$, hence in particular $U_x \subset \text{int}V = U$. Clearly $S \subset U \subset V$.