Proof that $\sqrt{2}$ is irrational is not convincing. Please help.

The second assumption can be avoided if it bothers you. You can say:

Suppose $\sqrt 2=\frac{a}{b}$ ($a,b$ not necessarily co-prime).

Let $c=\frac{a}{(a,b)}$ and $d=\frac{b}{(a,b)}$, where $(a,b)$ denotes the highest common factor of $a$ and $b$.

Then $\sqrt 2=\frac{c}{d}$, and $c$ and $d$ are co-prime.

Now proceed with the proof as above, with $c,d$ in place of $a,b$.

The first assumption implies the second. If a number is rational, then we can write it as $\frac ab$ with $a,b$ coprime. Thus if we cannot write a number as $\frac ab$ with $a,b$ coprime, then it is not rational.

This works since if $q$ is assumed to be rational, then it's always true that it can be expressed as a fraction of coprime integers. Thus if you disprove this, it means that $q$ is not rational.