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.
As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.
Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:
(1)If $\operatorname{Bd}(A)=\emptyset$
Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.
$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.
(2)If $A$ is closed and open.
A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$
But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$.
Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.
Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$
Best Answer
The first two characterisations (using $f^{-1}$) are derivable from each other : in any space $X$ we have the complements duality between interior and closure
$$\operatorname{int}(A)=X\setminus \overline{X\setminus A}$$
and $$\overline{A}= X\setminus \operatorname{int}(X\setminus A)$$
and moreover, $f^{-1}$ plays nice with complements, i.e.
$$f^{-1}[Y\setminus A]=X\setminus f^{-1}[A]$$
and complements reverse inclusions.
But we don't have the complement property for forward images (we do for injective functions, but generally $f[X\setminus A]\neq f[X]\setminus f[A]$ etc.)
The property that $f[\operatorname{int}(A)] \subseteq \operatorname{int}(f[A])$ characterises open maps, as can be easily seen. But the reverse
$$f[\operatorname{int}(A)] \supseteq \operatorname{int}(f[A])$$
does not characterise continuity, as can be seen from the examples here, (both implications with continuity fail) which I won't repeat here.