To begin with, here is an example in $\mathbb{R}$. Take $A=[0,1)$. Then the interior of $A$ is $(0,1)$, the interior of $A'$ is $(-\infty,0)\cup(1,+\infty)$, and the boundary is $\{0,1\}$. Note that this gives a partition of $\mathbb{R}$. If you take $A$ to be a line in the plane, then the interior of $A$ is empty, the interior of $A'$ is $A'$, and the boundary is the line $A$. Note this is again a partition of the plane.
As pointed out by AlexBecker, we need to be careful with the wording. I suppose you regard $\chi_A$ as a function from $\mathbb{R}$ to $\mathbb{R}$, equipped with the usual topology inherited from the norm/absolute value.
If a function is continuous at every point of $S\subseteq X$, then it is continuous on $S$ (good exercise on the induced topology). The converse is not true in general. For instance, $\chi_\mathbb{Z}$ is continuous on $\mathbb{Z}$, but it is discontinuous at every integer as a function on $\mathbb{R}$.
If a function is constant on $S\subseteq X$, it is easily seen to be continuous on $S$. So $\chi_A$ is continuous on the interior of $A$, and on the interior of $A'$, because it is constant there.
If a function is continuous on an open set $S\subseteq X$, then it is continuous at every point of $S$ (good exercise again). So $\chi_A$ is also continuous at every point of the interiors of $A$ and $A'$.
Now what about the boundary?
Assume that $\chi_A$ is continuous at some point $x$ of the boundary. You can use sequences. By assumption, there exist $(x_n)$ in $A$ and $(x'_n)$ in $A'$ which both converge to $x$. What is the value of $\chi_A$ on these sequences? What are these values supposed to converge to? Look for the contradiction.
Now you've shown that $\chi_A$ is discontinuous at every point of the boundary. But this does not mean it can't be continuous on the boundary.
Let $A\subset \Bbb R$.
Let $a\in A^o$.
Then $\exists$ a neighborhood $N_a$ of $a$ such that $N_a\subset A$.
Take an interval $(a-\epsilon, a+\epsilon) \subset \Bbb N_a$.
Define $a_n=a+1/n$.
For $n\gt 1/\epsilon$, we have $ a_n\in A$.
And clearly $a_n\to a$.
And $a_n\neq a,\forall n$.
$\therefore a\in A'$.
Best Answer
If $x\in\mathrm{int}(A\cap B)$, then there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq A\cap B$. And since $A\cap B\subseteq A$ and $A\cap B\subseteq B$, then...
If $x\in\mathrm{int}(A)\cap\mathrm{int}(B)$, then there exists $\epsilon_1\gt 0$ such that $(x-\epsilon_1,x+\epsilon_1)\subseteq A$, and there exists $\epsilon_2\gt 0$ such that $(x-\epsilon_2,x+\epsilon_2)\subseteq B$. Can you find a single $\epsilon$ that works for both sets? Then what can you say about $(x-\epsilon,x+\epsilon)$?