Let $U$ be a bounded, non-empty open subset of $\Bbb R$. For $x,y\in\Bbb R$ let
$$I(x,y)=\begin{cases}
[x,y],&\text{if }x\le y\\
[y,x],&\text{if }y\le x\;.
\end{cases}$$
$I(x,y)$ is just the closed interval with endpoints $x$ and $y$, irrespective of whether $x\le y$ or $y\le x$.
Define a relation $\sim$ on $U$ as follows: for $x,y\in U$, $x\sim y$ iff $I(x,y)\subseteq U$. It’s easy to check that $\sim$ is an equivalence relation. Let $J$ be a $\sim$-equivalence class. $J$ is bounded, so let $a=\inf J$ and $b=\sup J$. Note that $I(x,y)\subseteq J$ for any $x,y\in J$, so $J$ has no ‘holes’ and must be one of the four intervals with endpoints $a$ and $b$, $(a,b),[a,b),(a,b]$, or $[a,b]$.
Suppose that $a\in J$; then $a\in U$, so there is an open interval $(a-\epsilon,a+\epsilon)\subseteq U$. Let $u=a-\frac{\epsilon}2$; clearly $u\sim a\in J$, so $u\in J$, contradicting the choice of $a$. Thus, $a\notin J$. A similar argument shows that $b\notin J$. Thus, we must have $J=(a,b)$
That is, each of the $\sim$-equivalence classes is an open interval. The equivalence classes of any equivalence relation partition the underlying set of that relation, so the $\sim$-equivalence classes partition $U$ into disjoint open intervals.
The same argument works even if $U$ is unbounded, though in that cases one or two of the equivalence classes can be an unbounded open interval: if $U=\Bbb R$ there is just one, $\Bbb R$ itself, and otherwise you can get one interval of the form $(\leftarrow,a)$ or one interval of the form $(a,\to)$ or both.
You need $\Bbb Q$ only to show that this decomposition has only countably many pieces. And you do need $\Bbb Q$ or some other countable dense subset of $\Bbb R$ for that: the argument that I used above actually shows that in every linearly ordered space every open set can be written as a union of disjoint intervals, but there are non-separable linear orders in which some open sets have only decompositions with uncountably many pieces.
Best Answer
The answer is no, since an intersection of intervals is also an interval. Thus, if a closed set were to be a countable intersection of open intervals, it would have to be a closed interval, but there are closed sets that are not intervals.
For example (as @AlbertoTakase mentions in the comments below), consider the set $\{0\} \cup \{1\}$. This is a closed set since finite subsets of $\Bbb{R}$ are closed, but it is clearly not a closed interval. Hence, it cannot be written as a countable intersection of open intervals.