Prove that the function : $f(x) = x$ when $x$ is rational and $f(x) = 1 - x$ when $x$ is irrational is continuous at $x = \frac{1}{2}$
To prove the continuity it is necessary to prove that for every $\epsilon$ there exists a $\delta$ such that $|f(x) - f(\frac{1}{2})| < \epsilon$ whenever $|x -c| < \delta$ $- (1)$ Now if we choose $a \in (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$ and also $a $ is rational and choose $\epsilon < |a - \frac{1}{2}|$ then condition in $(1)$ is not satisfied. How to move from here?
You can't choose $\epsilon$ after choosing $\delta$. A $\delta$-$\epsilon$ proof is like a game between two players. Player A chooses an $\epsilon$, and player $B$ finds a $\delta$ so that the $f(x)$ stays within $\epsilon$ of the $f(a)$ whenever $x$ is within $\delta$ of $a$. If the function is continuous, then Player A can choose smaller and smaller $\epsilon$'s, and Player B will always be able to find a $\delta$ that works. If the function is not continuous, then Player A can find an $\epsilon$ so that no choice of $\delta$ works for Player B.
In your attempted proof Player A has "cheated" by shrinking their $\epsilon$ after $\delta$ was chosen. Instead, you should start with a fixed $\epsilon>0$, and then show that there is a $\delta>0$ so that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.