A boundary point of a set $A$ cannot be an interior point, but that’s not a problem: the theorem is that the closure of $A$ consists of those points that are in the interior of $A$ together with those that are in the boundary of $A$. In symbols,
$$\operatorname{cl}A=\operatorname{int}A\cup\operatorname{bdry}A\;,\tag{1}$$
and in fact the two sets on the righthand side of $(1)$ are disjoint. (By the way, a closed set need not have any boundary points at all: in $\Bbb R$ the only examples of this phenomenon are the closed sets $\varnothing$ and $\Bbb R$, but in more general topological spaces there can be many sets that are simultaneously open and closed and which therefore have empty boundary.)
You know that $x\in\operatorname{int}A$ if and only if $x$ has an open nbhd contained in $A$, and that $x\in\operatorname{bdry}A$ if and only if every open nbhd of $x$ intersects both $A$ and $\Bbb R\setminus A$. To prove $(1)$, I suggest that you first prove that
$\qquad\qquad\qquad\quad x\in\operatorname{cl}A$ if and only if for each open nbhd $V$ of $x$, $V\cap A\ne\varnothing$.
Once you have this, $(1)$ is pretty straightforward.
Your interpretation of the set at the end of your post is correct. The key to solving this problem is:
$\star$any open interval centered at a point will contain both rational and irrational numbers.
Isolated points: By $\star$ the points in $[0,3]$ are not isolated. The set of isolated points is $\mathbb{N} - \{1,2,3\}$. To show this, just take $\delta = 1/2$.
Limit points: No point in $\mathbb{N} - \{1,2,3\}$ can be a limit point since these points are isolated. By $\star$ the set of limit points is $[0,3]$.
Interior points: In the interval $(0,2)$, only rational numbers are in $A$, and in the interval $(1,3)$, only irrational numbers are in $A$. Since the points in $\mathbb{N} - \{1,2,3\}$ are isolated they cannot be interior. So the set of interior points of $A$ is
$$A_0 = (0,2) \cap (1,3) = (1,2).$$
Boundary points: Since all the points in $(1,2)$ are interior, they cannot be boundary points. Use $\star$ and the fact that the points in $\mathbb{N} - \{1,2,3\}$ are isolated to show that the set of boundary points is
$$\partial A = ([0,1] \cap \mathbb{Q}) \cup ([2,3] \cap \mathbb{R} - \mathbb{Q}) \cup \mathbb{N} = A - (1,2).$$
Best Answer
Every convergent sequence of elements of $A$ converges to an element of $A$ or to a natural number. Therefore, $\overline A=A\cup\mathbb N$. And $\mathring A=\emptyset$, since $A$ contains no interval. So, $\partial A=\overline A\setminus\mathring A=A\cup\mathbb N$. Finally, yes, the set of limit points of $A$ is $\mathbb N$.
Things are simpler with respect to the discrete metric, since then every set is both closed an open. So, $\overline A=\mathring A=A$. In particular, $\delta A=\emptyset$. And in a discrete metric space, the set of limit points of any subset is empty.