Prove that the product of a non-zero rational and irrational number is irrational. Could you please confirm if this proof is correct?

Theorem: If $q \neq 0$ is rational and $y$ is irrational, then $qy$ is irrational.

Proof: Proof by contradiction, we assume that $qy$ is rational. Therefore $qy=\frac{a}{b}$ for integers $a$, $b \neq 0$. Since $q$ is rational, we have $\frac{x}{z}y=\frac{a}{b}$ for integers $x \neq 0$, $z \neq 0$. Therefore, $xy = a$, and $y=\frac{a}{x}$. Since both $a$ and $x$ are integers, $y$ is rational, leading to a contradiction.
As I mention here frequently, this ubiquitous property is simply an instance of complementary view of the subgroup property, i.e.

THEOREM  $\  $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ 
comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where  $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$

Instances of this are ubiquitous in concrete number systems, e.g.
a is irrational, whereas b is rational.(both > 0)

Q: does the multiplication of a and b result in a rational or irrational number?:

Proof:

because b is rational:
    b = u/j where u and j are integers

Assume ab is rational:
ab = k/n, where k and n are integers.
a = k/bn
a = k/(n(u/j))
a = jk/un

before we declared a as irrational, but now it is rational; a contradiction. Therefore ab must be irrational.