Proving that if two integers have opposite parity, then their product is even

Yes of course that's correct.

We can also observe that

• if $a\in \mathbb{N}$ is even $\implies 2\,|\,a\,$ and $\,\forall b \in \mathbb{N} \quad 2\,|\,ab \implies ab$ is even.

Euclid's Lemma asserts that if a prime number $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ and $b$.

Since $2$ is a prime number, we can let $p = 2$. We now want to show that for integers $a$ and $b$ that have an opposite parity, $2$ must divide the product $ab$.

When two numbers have an opposite parity, then one of those numbers is even and the remaining number is odd. The definition of an even number is such that the number is divisible by $2$. The contrary is the definition of an odd number. This is why we let $p = 2$ in the first place.

Since either $a$ or $b$ must be even, the remaining integer consequently being odd, then $2$ must divide either $a$ or $b$. Therefore, $2$ must also divide $ab$ which makes that product an even number.

Go here for a proof of Euclid's Lemma.