If $4q^2 = p^2$, $4$ is a factor of $p^2$, but it does not follow that $4$ is a factor of $p$ only that $2$ is a factor of $p$. For example, $4$ is a factor of $36$, but $4$ is not a factor of $6$ (but $2$ is).
In general, if $a$ is prime and $a$ is a factor of $pq$ then $a$ must be a factor of $p$ or a factor of $q$ (or both); in particular, if $a$ is a factor of $p^2$, then $a$ must be a factor of $p$. It then follows that if $m$ is a product of distinct primes, then if $m$ is a factor of $p^2$, $m$ must be a factor of $p$. So the method used to show that the square root of a prime number is irrational extends to numbers which are the product of distinct primes, but no further.
Yes, you could instead (uniquely!) write $\,p = 2^j a\,$ and $\,q = 2^k b\,$ for $\,a,b\,$ odd and then deduce a contradiction by comparing the number of factors of $2,\,$ viz. $\,p^2\,$ has an even number of $2$'s but $\,2q^2\,$ has an odd number of $2$'s, a contradiction. This method uses only the very easily-proved existence and uniqueness of $2$-factorizations, i.e. representations of the form $\,2^j n,\,$ with $\,n\,$ odd (versus the much more powerful, and much more difficult to prove Fundamental Theorem of Arithmetic = existence and uniqueness of arbitrary prime factorizations).
As for your second question, yes, every fraction $\,A/B\,$ can be written with coprime numerator and denominator $\,a/b\,$ simply by cancelling their gcd $\,c,\,$ i.e. $\,A/B = ca/cb = a/b.\,$ By the maximality ("greatest") property of the gcd it follows that $\,d =\gcd(a,b) = 1,\,$ else $\,cd\,$ would be a common divisor of $\,A,B\,$ larger than the greatest common divisor $\,c=\gcd(A,B),\,$ contradiction. However, as above, it suffices to cancel only common factors of $\,2,\,$ so no knowledge of gcds is required.
Best Answer
It is, but I think you need to be a little bit more careful when explaining why $5$ divides $p^2$ implies $5$ divides $p$. If $4$ divides $p^2$ does $4$ necessarily divide $p$?