Is the following method for proving density of irrational numbers in real numbers without using rational numbers density in real numbers rigorous?

The motivation for this question is:

I told my friend to use:

$\forall x_{1}, x_{2} \in \mathbb{R}, x_{1} < x_{2}, \exists r \in \mathbb{Q}: x_{1} < r <x_{2}.$

To prove:

$\forall x_{1}, x_{2} \in \mathbb{R}, x_{1} < x_{2}, \exists s \in \mathbb{Q^{c}}: x_{1} < s <x_{2}.$

But what surprised me is: He proved it without using the density of rational numbers in $\mathbb{Q}$ as following:

By contradiction:

Assume $\exists x_{1}, x_{2} \in \mathbb{R}, x_{1} < x_{2}, \forall s \in \mathbb{Q^{c}}: s \notin(x_{1},x_{2}).$

Thus $(x_{1},x_{2}) \subset \mathbb{Q}$, this leads to the contradiction, because $\mathbb{Q}$ is countable infinite, where $(x_{1},x_{2})$ is uncountable.

Is this method good and rigrous to prove it?

My reason for checking is that this method may be circular without realizing that or it may depend on density of rationals implicitly, so that I want your opinions.