Square root of $2$ is irrational

Before you continue the argument you write the fraction $m/n$ in lowest terms. Then the numerator and denominator can't both be even.

we can assume that $$\gcd(m,n)=1$$ and $$2=\frac{m^2}{n^2}$$ then $$2n^2=m^2$$ thus the left-hand side is even and so $$m^2$$ this is a contradiction, both numbers $m,n$ can not be even

Because if $n=2n_1$ and $m=2m_1$ then $\frac{m_1^2}{n_1^2}=2$... and we get an infinite series $n>n_1>n_2>...$ of natural numbers, which is impossible.