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.
This sometimes called the Popcorn Function, since if you look at a picture of the graph, it looks like kernels of popcorn popping. (Also called the Thomae's Function.)
To prove it is discontinuous at any rational point, you could argue in the following way: Since the irrationals that are in $(0,1)$ are dense in $(0,1)$, given any rational number $p/q$ in $(0,1)$, we know $f(p/q) = 1/q$. But let $a_{n}$ be a sequence of irrationals converging to $p/q$ (by the definition of density). Then $\lim \limits_{a_{n} \to (p/q)} f(x) = 0 \neq f(p/q)$. Thus, $f$ fails to be continuous at $x = p/q$.
Now, to prove it is continuous at every irrational point, I recommend you do it in the following way:
Prove that for each irrational number $x \in (0,1)$, given $N \in \Bbb N$, we can find $\delta_{N} > 0$ so that the rational numbers in $(x - \delta_{N}, x + \delta_{N})$ all have denominator larger than $N$.
Once you have the result from above, we can use the $\epsilon-\delta$ definition of continuity to prove $f$ is continuous at each irrational point.
So, first prove 1. (It's not too hard.) Once you've done that, you can accomplish 2. in the following way:
Let $\epsilon > 0$. Let $x \in (0,1)$ be irrational. Choose $N$ so that $\frac{1}{n} \leq \epsilon$ for every $n \geq N$ (by the archimedian property).
By what you proved in 1., find $\delta_{N} > 0$ so that $(x - \delta_{N}, x + \delta_{N})$ contains rational numbers only with denominators larger than $N$.
Then if $y \in (x-\delta_{N}, x + \delta_{N})$, $y$ could either be rational or irrational. If $y$ is irrational, we have $|f(x) - f(y)| = |0 - 0| = 0 < \epsilon$ (since $f(z) = 0$ if $z$ is irrational).
On the other hand, if $y$ is rational, we have $y = p/q$ with $q \geq N$. So $|f(x) - f(y)| = |0 - f(y)| = |f(y)| = |1/q| \leq |1/N| < \epsilon$, and this is true for every $y \in (x-\delta_{N}, x + \delta_{N})$.
Thus, given any $\epsilon > 0$, if $x \in (0,1)$ is irrational, we found $\delta > 0$ (which was actually $\delta_{N}$ in the proof) so that $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.
Best Answer
The idea that "a function is continuous if (and only if) its graph can be drawn without lifting one's pen(cil)" is sometimes adequate for communicating with non-mathematicians, but is technically flawed for multiple reasons.
For convenience, let's call this "condition" pen continuity.
First, as other answers note, a function must be continuous on an interval to have any hope of being pen continuous. Unfortunately for "pen continuity", there are a couple of reasons a function might be continuous (to a mathematician, using the $\varepsilon$-$\delta$ definition), but not continuous on an interval:
A function can be continuous at a single point (such as the function in your post), or at each point of a complicated set that contains no interval of real numbers (such as Thomae's function, which is continuous at $x$ if and only if $x$ is irrational).
A function can be continuous at every point of its domain, but the domain is not an interval (and perhaps contains no interval). Think, for example, of Przemysław Scherwentke's example $f(x) = 1/x$ for $x \neq 0$, which is continuous throughout its domain (the set of non-zero real numbers), or of the zero function defined on an arbitrary set of real numbers (which can be nastier than the human mind can comprehend).
So, let's focus on (real-valued) functions that are continuous at every point of an interval. Depending on your definition of a pen, not every continuous function is pen continuous (!). If a "pen" is a mathematical point, and "draw" has its ordinary meaning ("the pen can be traced along the graph in finite time", say), then most continuous functions are not pen continuous, because their graphs have infinite length over arbitrary subintervals (or "are not locally rectifiable", in technical terms). (The Koch snowflake curve isn't a graph, but may be a familiar non-rectifiable example.)
To emphasize, a "typical" continuous function is nowhere-differentiable: Its graph looks something like an EKG or a seismograph tracing or the curves you draw after drinking 50 cups of espresso. "Zooming in" only reveals details at smaller and smaller scales, peaks and valleys whose total length (over an arbitrarily short subinterval of the domain) may well be infinite. Only functions of bounded variation have graphs of finite length, and that's a "thin" subset of all continuous functions.
[If instead you want to think of "real" pens, whose tip has positive radius, you arrive at mathematically interesting territory, including Hausdorff measure and geometric measure theory.]
The bottom line (literally!) is, a mathematician mustn't conflate "continuity" with "pen continuity".