Closed set $F$ is the boundary of any subset of $\mathbb{R}^n$ I need show that any closed subset $F\subset\mathbb{R}^n$ is the boundary of some set $A$ in $\mathbb{R}^n$.
Intuition tells me to take $A=F\setminus(\mathbb{Q}^n\cap int(F))$ and $int(F)$ is the set of interior points of $F$ but I can't prove that boundary$(A)\subset(F)$
 A: Since $A \subset F$ you have 
$$\partial A \subset \overline{A} \subset \overline{F} =F$$
where $\partial A$ denotes the boundary of $A$ and $\overline{X}$ denotes the closure of $X$.
P.S. If it is the other implication you cannot prove, here is how you can prove it:
If $x \in F$ then we have two possibilities:
case 1: $x \in \text{Int}(F)$
In this case $B_r(x) \subset F$ for some $r$ and then
$$
B_r(x) \cap \mathbb{Q} \subset A \\
B_r(x) \cap (\mathbb{R}\setminus\mathbb{Q}) \cap A =\emptyset 
$$
It is easy to show from here that $x$ is a boundary point of $A$.
case 2: $x \notin \text{Int}(F)$.
Then we have $x \in A$. Also as $A$ has no interior we have $\text{Int}(A)=\emptyset$ and hence
$$A \subset \overline{A} = \overline{A}\setminus\text{Int}(A)=\partial A$$
This shows that 
$$A \subset \partial A$$
and hence, as $x \in A$ we get $x \in \partial A$ .
In both cases we showed that $x \in F \Rightarrow x \in \partial A$.
A: This answer shows that if a space $X$ can be written as the union of two disjoint dense subsets (such a space is called resolvable), then every closed set in it is the boundary of some other set.
This certainly holds for all $\mathbb{R}^n$ (take all rational points, i.e. $\mathbb{Q}^n$ and its complement, e.g.).
The idea is quite general.
A: If x is in the boundary of A the it is an adherent point of A and thus an adherent point of F.  F is closed so all adherent points of F are in F.  So x is in F.
A: Your problem is, that $\mathbb{Q}^n\cap F$ might be not dense or even empty. Instead of the section with $\mathbb{Q}^n$ for $k\in\mathbb{N}$ take a set $N_{k}$ of nodes inside your set $F$ wich are at least $\frac{1}{2k}$ away from each other and for which each point of your set $F$ is maximally $\frac{1}{k}$ away from one of the nodes in $N_k$. Then take for $A$ the union $A:=\cup_{k}N_k$ and see that the boundary of that set is the closure of your original set $F$. (since the interior is empty and the closure is the closure of your set; since $F$ is closed it equals its closure; by construction of $N_k$ you can easily choose sequences converging to Points in the closure of $F$)
So the idea is to construct a dense, at most countable subset.
