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The Cantor set $C$ is perfect, hence all it points are limit points. Then for every $x\in C$ exists at least a sequence such that $(x_n)\to x$. Then it make me think that the definition of (dis)continuity (at any point $x$) can be applied here, that is

$$\forall \epsilon>0,\exists \delta>0:|y-x|<\delta\implies |f(x)-f(y)|<\epsilon$$

But, at the same time, by the induced topology from $\Bbb R$ is clear that any subset of $C$ is open and closed, hence $f$ is trivially continuous. Can someone clarify this question? Thank you.

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    The fundamental intervals that make up the Cantor set are clopen, yes, but I don't think that every set is both open and closed. For example, $\{1/3^n:n\in\mathbb Z\}$ is neither open nor closed. – 2017-02-08
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    Not every subset is open and closed. The Cantor set is not discrete. It is just $\textit{really}$ disconnected. – 2017-02-08
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    Also, any space with at least two points can't be perfect if it is a discrete space. – 2017-02-08
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    @MarkMcClure oh, I see... I supposed that it topology will be "simple". – 2017-02-08
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    In a sense it's close to discrete because it's homeomorphic to the Cartesian product of countably many discrete spaces, but in another sense it's as far from discrete as you can get because every point is a limit point. – 2017-02-08

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Let $c\in C$ be a point in the Cantor set and define $f:C\to\mathbb R$ by $$f(x)=\left\{\begin{array}{ll}0&\mbox{ if }x=c\\2&\mbox{ otherwise }\end{array}\right .$$ Then $f^{-1}((-\infty,1))=\{c\}$, and since $c$ is a limit point of $C$ it cannot be open. Thus $f$ is not continuous.

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    I dont follow the example... the set $(-\infty,1)$ does not belong to the image of $f$. With your function we have that $f(C)=\{0,2\}$ what is closed, but $f^{-1}(0)=C\setminus\{c\}$ what is not closed, so yes, your example works but for a different reason. – 2017-02-08
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    @Masacroso Even if an open set is not contained in the image of a function, its inverse image must be open for the function to be continuous. – 2017-02-08