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.
Hint. Let's $\tau$ is the topology of your question. You can think of a more concrete way of explaining the face of this open topology.If $O\in \tau$ and $O\neq \emptyset$ then $O^c$ is finite. That is, thare is $n$ numbers $x_1<x_2<\ldots, x_{n-1}<x_n$ such that $O^c=\{x_1, \ldots, x_n\}$. Then,
$$
O=(-\infty,x_1)\cup( x_1,x_2)\cup\ldots\cup( x_{n-1},x_n)\cup(x_n,\infty)
$$
This means that each nonempty open $O\in\tau$ contains at least two intervals of infinite type
$$
(-\infty,a) \mbox{ and } (b,+\infty) \mbox{ with } a\leq b.
$$
Therefore the only open that can be contained in the set S is empty. Since, by definition, the interior of a set $S$ is the union of all open contained in $S$ then the interior of $S$ is the empty set ($S$ contains only empty set).
Best Answer
To clarify:
The boundary is the set of points whose neighbourhoods always intersect $A$, but do not lie in the interior of $A$.
Its interior is the set of points who have a neighbourhood which lies entirely in the boundary. As boundary points lie in the closure, this neighbourhood must also intersect $A$.
The closure of the interior of the boundary are the points for which every neighbourhood intersects the interior of the boundary.
Now assume $a \in cl(int(\partial A ))$. Let $U_a$ be one of its neighbourhoods; it intersects $int(\partial A)$ in at least one point $y$. Let $V_y$ be a neighbourhood of $y$ which lies completely in the interior boundary. The intersection $W_y := U_a \cap V_y$ is also a neighbourhood of $y$ which lies completely in the interior boundary; furthermore, it is a subset of $U_a$. As argued before, $W_y$ must contain a point $z$ of $A$. This point $z$ is also a point of the interior boundary. Therefore, $z \in A \cap int(\partial A)$.
We have just proven that every neighbourhood of $a$ intersects a point $z \in A \cap int(\partial A)$. By definition it then follows that $$cl(int(\partial A )) \subset cl( A \cap int(\partial A) )$$
which concludes the proof.