Suppose $F\subset V\subsetneq \mathbb {R}$ where $F$ is closed and $V$ is open. I want to show that $\exists$ an open set $U\subset \mathbb {R}$ satisfying $F\subset U\subset \overline {U}\subset V$. My proof is below, but I think it's problematic since I never use the condition that $F$ is closed.


For each $x\in F$, pick $r\_{x}>0$ such that $(x-r\_{x},x+r\_{x})\subset V$. Then define $U=\bigcup\{(x-\frac {r\_{x}}{2},x+\frac {r\_{x}}{2})\}\subset V$. Then in fact, $F\subset U$. This is because countable union of open sets is open and $U$ is an open cover of $F$. It's trivial that $U\subset \overline{U}$. Now, what remains is to show that $\overline{U}\subset V$. First of all, $U\subset V$ since $U$ is just a collection of open intervals, and each one of them is in $V$. In fact, under this construction, $U\subsetneq V$. Therefore, for $y\in \overline{U}\setminus U$, $\exists u\in U$ such that $y\in B(u,r\_{u})$. Take $r\_{y}=r\_{u}-d(y,u)$. Then $B(y,r\_{y})$ is an open ball inside of $V$. Thus, $y\in V$ and $\overline{U}\subset V$.



Between an open set and its closed subset
$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\de}{\delta}\newcommand{\ol}{\overline}$The property of $\R$ that you want to prove is that the topological space $\R$ is [normal](https://en.wikipedia.org/wiki/Normal_space). [All metric spaces are perfectly normal](https://en.wikipedia.org/wiki/Normal_space#Examples_of_normal_spaces) and hence normal. This proves your desired result.


That all metric spaces are normal immediately follows e.g. from Theorems 5.1.3 and 5.1.5 of the book by [Engelking](https://rads.stackoverflow.com/amzn/click/com/3885380064): Theorem 5.1.3 (due to Stone; see also the short proof by [M. E. Rudin](https://www.jstor.org/stable/2035708)) states that every metrizable space is [paracompact](https://en.wikipedia.org/wiki/Paracompact_space), and Theorem 5.1.5 (due to Dieudonné) states that every paracompact space is normal.




---


Here is an elementary, self-contained proof:


For each $n\in\Z$, the set
\begin{equation\*}
 F\_n:=F\cap[n,n+1] \tag{1}
\end{equation\*}
is closed and bounded, and hence a compact subset of the open set $V$, so that there is some $\de\_n\in(0,1)$ such that
\begin{equation\*}
 \ol{U\_n}\subseteq V, \tag{2}
\end{equation\*}
where
\begin{equation\*}
 U\_n:=(F\_n)\_{\de\_n}, \tag{3}
\end{equation\*}
the open $\de\_n$-neighborhood of $F\_n$.


Let finally
\begin{equation\*}
 U:=\bigcup\_{n\in Z}U\_n, \tag{4}
\end{equation\*}
so that $U$ is an open set containing $F$.


It remains to show that $\ol U\subseteq V$. Take any $x\in\ol U$, so that $x=\lim\_{k\to\infty}x\_k$ for some sequence $(x\_k)$ in $U$. Without loss of generality, for some natural $N$ and all natural $k$ we have $x\_k\in U\cap[N-1,N+1]$.
The conditions $F\_n\subseteq[n,n+1]$, (3), and $\de\_n\in(0,1)$ imply
\begin{equation\*}
 \ol{U\_n}\subseteq[n-1,n+2]. 
\end{equation\*}
Hence, by (4), for all natural $k$ we have
\begin{equation\*}
x\_k\in\bigcup\_{n\colon\,[n-1,n+2]\cap[N-1,N+1]\ne\emptyset}U\_n\ 
\subseteq\bigcup\_{n=N-3}^{N+2}U\_n. 
\end{equation\*}
So,
\begin{equation\*}
 x\in\bigcup\_{n=N-3}^{N+2}\ol{U\_n}\subseteq V, 
\end{equation\*}
by (2). Thus, indeed $\ol U\subseteq V$. $\quad\Box$


(The part $V\ne\R$ of your condition $V\subsetneq\R$ is not needed here; looking at this condition, it seems that you are using $\subset$ for $\subseteq$.)


