2. ## Re: continuity

That argument would work fine to *suggest* that there is a discontinuity. However, it would not *prove* that there is a discontinuity. There are many continuous functions such that f(1.41)=1.41, f(1.414...)=0 and f(1.42)=1.42.

3. ## Re: continuity

The standard argument uses the observation that for any $\epsilon>0$, there are only finitely many points $a_i$ such that $f(a_i)>\epsilon$.

Spoiler:
So given an irrational point $x$, let $\delta=\min\{|x-a_1|,|x-a_2|,...,|x-a_n|\}$. Then if $|x-y|<\delta$, $|f(y)-f(x)|=f(y)<\epsilon$, so $f$ is continuous at the irrational point $x$.

It's straightforward to show $f$ is discontinuous at any rational point just using the fact that any interval contains an irrational number.

Spoiler:
If $x\in\mathbb{Q}$, then $f(x)=c>0$, so letting $\epsilon=c/2$ shows that since any small interval around $x$ contains an irrational point $y$, $|f(x)-f(y)|=c > c/2$.