If \(\sqrt{2}\) is rational, then there are two integers \(a\) and \(b\) such that:
$$\begin{align} \sqrt{2} &= \frac{a}{b} \\ b\sqrt{2} &= a \end{align}$$
This means there exists integers \(b\) such that \(b\sqrt{2}\) is an integer. Let \(S\) be a set that contains these integers:
$$ S = \{ b\sqrt{2} \quad | \quad b \text{ and } b\sqrt{2} \text{ are positive integers}\}$$
According to the well-ordering property, \(S\) contains a smallest element \(s\). Let \(s=t\sqrt{2}\). This means:
$$\begin{gather} s\sqrt{2} = 2t \\ (t \in \mathbb{N}) \wedge (s \in \mathbb{N}) \implies 2t - s \in \mathbb{Z} \end{gather}$$
We can rewrite this as:
$$\begin{gather} 2t - s = \\ s\sqrt{2} - s = s\sqrt{2} - t\sqrt{2} = \\ (s-t)\sqrt{2} \in \mathbb{Z} \end{gather}$$
Since \([s = t\sqrt{2}]\) and \([\sqrt{2} \gt 1]\), then \(s \gt t\). Which means \([s-t]\) is a positive integer. This also means that \([(s-t)\sqrt{2}]\) is a positive integer. By definition of \(S\), \([(s-t)\sqrt{2}]\) is also in \(S\).
Also:
$$\begin{gather} \sqrt{2} \lt 2 \\ s\sqrt{2} \lt 2s \\ s\sqrt{2} - s \lt s \\ s\sqrt{2} - t\sqrt{2} \lt s \\ \therefore (s-t)\sqrt{2} \lt s \end{gather}$$
Since \([(s-t)\sqrt{2}]\) exists in \(S\) and \([(s-t)\sqrt{2} \lt s]\), then we have a contradiction.