Did you want the proof to be from first principles?

The following proof by contradiction assumes some basic knowledge, such as "any rational can be written in lowest terms" and "the square of an odd integer is odd".

Suppose $x$ is a rational such that $x^2= 2$, then $x$ can be written in lowest terms as $p/q$,

where $p$ and $q$ are integers and $\gcd(p,q) = 1$.

As $x^2 = 2$ implies $(p/q)^2 = 2$, $p^2 = 2q^2$, which is even.

As the square of an odd integer is odd, $p$ is even, so $p = 2m$, where $m$ is an integer.

Hence $q^2 = 2q^2/2 = p^2/2 = (2m)^2/2 = 2m^2$, which is even.

As the square of an odd integer is odd, $q$ is even.

As $p$ and $q$ are both even, $\gcd(p,q) \geqslant 2$, which contradicts $\gcd(p,q) = 1$.

Hence the supposition that $x$ is a rational such that $x^2 = 2$ is false.