Suppose $\sqrt 2$ were rational. Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.
Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.
Squaring both sides,
$2 = \frac 21 = \frac{a^2}{b^2}$.
Lowest terms representations of rational numbers are unique, so we have $a^2 = 2$ and $b^2=1$.
But there is no such integer $a$, and therefore we have a contradiction and $\sqrt 2$ is irrational.
I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound.
Best Answer
Your proof is correct. Quicker, from $\rm\:\color{#0a0}{(a,b)\!=\!1},\ 2\:\!b^2 = a^2\:$ we deduce by Euclid's Lemma ($\color{#c00}{\rm EL}$) that $\rm\:\color{#0a0}{b\:|\:a}\:\!a\:\Rightarrow\:b\:|\:a\,\ $ so $\rm\:\sqrt{2} = a/b\in \mathbb Z.\:$
In your proof, the only "uniqueness" result we need on reduced fractions is the following:
$\rm{\bf Lemma}\ \ (a,b)\!=\!1,\ \dfrac{a}b = \dfrac{A}B\:\Rightarrow\:b\:|\:B\ \ $ Proof $\rm\ \ Ab=aB\:\Rightarrow\:b\:|\:aB\:\Rightarrow\:b\:|\:B\:$ by $\rm\color{#c00}{EL}.\ \small\bf QED$
$\rm Hence\quad\ \, (a,b)\!=\!1,\ \dfrac{a}b = \dfrac{2b}{a}\ \Rightarrow\:b\:|\:a\ \Rightarrow\ \sqrt{2}=\dfrac{a}b\in\mathbb Z.\ $ So you don't need $\rm\:(a^2,b^2)=1.$
This method of proof generalizes to arbitrary square roots, e.g. see here.
The converse of the Lemma is also true, i.e. Euclid's Lemma is equivalent to this uniqueness property of reduced fractions. For more on this topic see my posts on unique fractionization.