Let $g(x)$ be:
\begin{cases} 0 \space\space \text{if}\space\space x \in \mathbb Q \\ x\space\space \text{if} \space\space x \in \mathbb R - \mathbb Q \\ \end{cases}
That's what I've tried since I started:
If $g(x)$ is continuous so $\lim\limits_{x\to a}g(x) = g(a)$. By $\varepsilon - \delta$ definition:
$\forall \varepsilon \gt 0 \space \exists \delta \gt 0 \space\space$ such that $\space\space 0 \lt|x - a| \lt \delta \implies |g(x) - g(a)| \lt \varepsilon$.
First case: suppose $a \in \mathbb Q$ and $x \in \mathbb R - \mathbb Q$, so:
$|g(x) - g(a)| = |x - 0| = |x| \lt \varepsilon$.
Second case: suppose $a \in \mathbb R - \mathbb Q$ and $x \in \mathbb Q$, so:
$|g(x) - g(a)| = |0 - a| = |-a| = |a| \lt \varepsilon$.
It doesn't matter if $x$ and $a$ are greater or less than $0$. Considering I set both to be greater than $0$, $|x|$ and $|-a|$ and equal to $x$ and $a$ respectively, and, by definition, I can set $\varepsilon$ to be any number, so I can make it less than $x$ and less than $a$ at the same time, so the definition wouldn't hold. If they're negative, $|x|$ = $-x$ and $|-a| = -a$, I can also let $\varepsilon$ be any number such that it will be less than $-x$ and $-a$ simultaneously.
The only case where the definition would hold is when I set $a = 0$ in the first case, so it would be:
Given $\varepsilon \gt 0$ $\space$ $\exists \delta \gt 0$ $\space\space$ such that $\space\space$ $0 \lt |x - a| \lt \delta \implies |g(x) - g(a)| \lt \varepsilon$.
If $a = 0$ and $x$ remains a irrational number:
$0 \lt |x| \lt \delta$ $\implies$ $|x| \lt \varepsilon$. Let $\delta = \varepsilon$ and we conclude that g is continuous at $0$.
Is it enough? I've seen some proofs to this before, and they always used the Squeeze Theorem, I wanted to do this way because the approach they did seemed like a bit harder to visualize at the beginning.