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I want to prove highlighted part by epsilon-delta method.

So by definition of continuity:Function $f$ is continuous at $a$ if for every $\epsilon >0$ there exists a $\delta >0$ such that $|x - a| < \delta => |f(x) - f(a)| < \epsilon$. And we must show, that this fact doesn't hold.

Let's pick $\epsilon = 1/2$

Suppose $a$ is irrational:$|x - a| < \delta => |f(x) - f(a)| < \epsilon$

From this follows $|f(x)| < \epsilon$ and if we will pick any rational number $x$ within $\delta$ and $\sigma$ then $|f(x)| = 1 > 1/2 = \epsilon$. So function is not continuous if a is irrational

The same strategy for $a$ is rational: $|x - a| < \delta => |f(x) - f(a)| < \epsilon$

From this follows $|f(x)| < \epsilon$ and if we will pick any irrational number $x$ within $\delta$ and $\sigma$ then $|f(x)-1| = 1 > 1/2 = \epsilon$. So function is not continuous if x is rational.

Is my proof correct? if not, what exactly is wrong? Or maybe I missed some important details?

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    More or less, but presentation is a little confused. You start by defining continuity at the point $a$, but then test it at the point $x$. Replace "Suppose x is irrational" with "Suppose $a$ is irrational". In the next paragraph replace $a$ with $x$ and "if we pick any rational number $a$" with "for any rational number $x$ within $\delta$ of $a$". Then similarly for $a$ rational. – 2017-02-15
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    @Paul or, you are completely right!Thank you:) – 2017-02-15
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    Those three questions suggest a fourth one : can we construct a function $f:\mathbb{R}\to\mathbb{R}$ such that $f$ is continuous only at rationals ? The answer is surprisingly "no" and this is a consequence of Baire's theorem (it can be shown that the set of continuity points of $f$ is a $G_\delta$, that is the intersection of a countable family of open subsets, and by Baire's theorem, $\mathbb{Q}$ is not a $G_\delta$). – 2017-02-15

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