Find the points where $f$ is continuous. 
Question: Let $f(x)$ be the function defined on the interval $(0,1)$ by $$f(x)=\left\{\begin{array}{l}x(1-x)\hspace{1.4 cm}\text{if $x$ is rational,}\\\frac{1}{4}-x(1-x)\hspace{0.5 cm}\text{if $x$ is not rational.}\end{array}\right.$$ Then find the points where $f$ is continuous. 

My approach: Select any rational point $a\in(0,1)$. We know that $\exists$ a sequence $\{a_n\}_{n\ge 1}$ of rational numbers such that $$\lim_{n\to\infty}a_n=a$$ and $\exists$ a sequence $\{b_n\}_{n\ge 1}$ of irrational numbers such that $$\lim_{n\to\infty}b_n=a.$$ 
Now if $f$ is continuous at $a$, then we must have $$\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}f(b_n)=f(a)\\\iff \lim_{n\to\infty}a_n(1-a_n)=\lim_{n\to\infty} \frac{1}{4}-b_n(1-b_n)=a(1-a)\\\iff a(1-a)=\frac{1}{4}-a(1-a)\\\iff 2a^2-2a+\frac{1}{4}=0\\\iff a=\frac{2\pm\sqrt 2}{4}.$$ But we have assumed $a$ to be rational so, $a\neq \frac{2\pm\sqrt 2}{4}$. Thus a contradiction is obtained. Thus $f$ is not continuous at any $a\in\mathbb{Q}$.
Now select any irrational point $b\in(0,1)$. Again, we know that $\exists$ a sequence $\{c_n\}_{n\ge 1}$ of rational numbers such that $$\lim_{n\to\infty}c_n=b$$ and $\exists$ a sequence $\{d_n\}_{n\ge 1}$ of irrational numbers such that $$\lim_{n\to\infty}d_n=b.$$ 
Now if $f$ is continuous at $b$, then we must have $$\lim_{n\to\infty}f(c_n)=\lim_{n\to\infty}f(d_n)=f(b)\\\iff \lim_{n\to\infty}c_n(1-c_n)=\lim_{n\to\infty} \frac{1}{4}-d_n(1-d_n)=\frac{1}{4}-b(1-b)\\\iff b(1-b)=\frac{1}{4}-b(1-b)\\\iff 2b^2-2b+\frac{1}{4}=0\\\iff b=\frac{2\pm\sqrt 2}{4}.$$
Thus if $f$ is continuous at some point $x\in(0,1)$, then $x=\frac{2\pm\sqrt 2}{4}.$ 
Now do we need to prove that the converse holds true, that is if $x=\frac{2\pm\sqrt 2}{4}$, then $f$ is continuous at $x$ to make the proof correct?
 A: You can shorten this by  considering $any$ $a\in (0,1).$ The 1st part of your argument shows that IF $f$ is continuous at $a$ THEN $a\in \{(2\pm \sqrt 2)/4\}.$
Now the functions $g(x)=x(1-x)$ and $h(x)=1/4-g(x)$ are continuous at all $x\in (0,1)$. And $a\in \{(2\pm \sqrt 2)/4\}\implies g(a)=h(a).$ So IF $a\in \{(2\pm \sqrt 2)/4\}$ THEN $f$ is continuous at $a.$
A: Please note that this is not a complete answer and I would like to complete it with some help.
Let us take any sequence $\{s_n\}_{n\ge 1}$ of real numbers such that $$\lim_{n\to\infty}s_n=\frac{2+\sqrt 2}{4}.$$ 
Also define $\{q_n\}_{n\ge 1}$ to be any arbitrary subsequence of $\{s_n\}_{n\ge 1}$ such that it comprises of only rational numbers, $\{r_n\}_{n\ge 1}$ to be any arbitrary subsequence of $\{s_n\}_{n\ge 1}$ such that it comprises of only irrational numbers and finally $\{t_n\}_{n\ge 1}$ to be any arbitrary subsequence of $\{s_n\}_{n\ge 1}$ such that it comprises of both rational and irrational numbers.
Now we know that a sequence $\{k_n\}_{n\ge 1}$ is convergent and converges to $k$ if and only if all of its subsequences are convergent and converges to $k$. $(*)$
Thus using $(*)$ we can conclude that, since $\{s_n\}_{n\ge 1}$ is convergent and converges to $\frac{2+\sqrt 2}{4}$, thus we can conclude that both $\{q_n\}_{n\ge 1}$ and $\{r_n\}_{n\ge 1}$ are convergent sequences and their limit is $\frac{2+\sqrt 2}{4}$.  
Now in order to prove that $f$ is continuous at $x=\frac{2+\sqrt 2}{4}$, it is both necessary and sufficient to prove that $$\lim_{n\to\infty}f(s_n)=f\left(\frac{2+\sqrt 2}{4}\right)=\frac{1}{8}\hspace{0.5cm}(**).$$
Now to prove that $(**)$ holds true it is again necessary and sufficient to prove that any subsequence of $\{f(s_n)\}_{n\ge 1}$ is convergent and converges to $\frac{2+\sqrt 2}{4}$. 
Now since $\{q_n\}_{n\ge 1}$ and $\{r_n\}_{n\ge 1}$ are convergent sequences, implies both the sequences $\{f(q_n)\}_{n\ge 1}$ and $\{f(r_n)\}_{n\ge 1}$ are convergent and we have $$\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}q_n(1-q_n)=\frac{1}{8}$$ and $$\lim_{n\to\infty}f(r_n)=\lim_{n\to\infty}\left(\frac{1}{4}-r_n(1-r_n)\right)=\frac{1}{8}.$$
Thus if we can prove that $$\lim_{n\to\infty}f(t_n)=\frac{1}{8},$$ we will be done. 
How to prove the same?
Also a similar analysis for $x=\frac{2-\sqrt 2}{4}$ will help us in proving that $f$ is continuous at $x=\frac{2-\sqrt 2}{4}.$ 
Hence, we will be able to conclude that $f$ is continuous only at $x=\frac{2\pm \sqrt2}{4}$. Observe that we can prove by contradiction that $f$ is not continuous at all other points.   
