Where does the proof of $sqrt 2$ is irrational break down when trying to prove the same for $sqrt 4$?

I'm trying find where the common proof by contradiction that $\sqrt 2$ is irrational breaks down when trying to prove $\sqrt 4$ is irrational.

Assume $\left(\frac pq\right)^2=4$ and $\gcd(p,q)=1$. I guess I could just let $p=2, q=1$ and be done, but why is that an adequate failure of the proof? If it's not, then where else does the proof break down?

Continuing the "proof": Then $p^2=4q^2$, so $p$ is even, that is, $p=2r$. That implies that $4r^2=4q^2$, or that $q=r$.

If $p=2r$ and $q=r$ then $\gcd(p,q)=r$ which contradicts $\gcd(p,q)=1$, leaving us with the untenable conclusion that $\sqrt 4$ is irrational.

Answer

It does not contradict the relatively prime assumption because we can take $r=1$.