Let $S \subset \mathbb{R^n}$ and let $f$ be a function from $S$ into $\mathbb{R^m}$. A function $f$ is continuous at $a$ if $\forall \ \epsilon > 0, \ \exists\ \delta > 0$ such that if $\bigg| x - a \bigg| < \delta \rightarrow \bigg|f(x) - f(a) \bigg| < \epsilon$ For
Prove that the indicator function: $$f(x) = \begin {cases} 0 & x\notin \mathbb{Q}\\ 1 & x \in \mathbb{Q} \end{cases} $$
is discontinuous at every point $a \in \mathbb{R}$.
My professor assigned this exercise as practice for us to do. Specifically he asked us to show "formally" that it is the case that the indicator function is discontinuous. I understand exactly why it is discontinuous, I would like to know if the manner in which I proved it is "mathematically rigorous".
Proof:
By the Archimedian properties of $\mathbb{R}$, for all $x,y \in \mathbb{Q}$, there exists $w \notin \mathbb{Q}$ such that $x < w < y$. As well for all $q_1, q_2 \notin \mathbb{Q}$, there exists $r \in \mathbb{Q}$ such that $q_1 < r < q_2$
With this being the case there are two cases to examine:
$x \in \mathbb{Q}, \ a \notin \mathbb{Q}$ AND $a \in \mathbb{Q}, \ x \notin \mathbb{Q}$
Without loss of generality let's suppose:
$x \in \mathbb{Q}, \ a \notin \mathbb{Q}$
The other follows directly.
Let $\epsilon = 0$, $\delta > 0$, and let $x = \frac{\delta}{2} + a$
Therefore by definition: $$ \bigg| x - a \bigg| = \bigg| (\frac{\delta}{2} + a) - a \bigg| = \frac{\delta}{2} < \delta$$
This means: $$\bigg|f(x) - f(a) \bigg| = |1 - 0| = 1 > \epsilon = 0$$
I feel that this is sufficient because by invoking the Archimedian properties I take care of the whole issue of the denseness with $\mathbb{Q}$. But this is why I am asking the question here. Could I get feedback?