proof writing - How do we know all numbers alternate between odd and even numbers?

Basically, I tried proving that multiplying an odd and even number together gives you an even number to a friend of mine. This is what I said.

Let's first take a random number $k$. It doesn't matter if its odd or even. If we then multiply this by $2$, we get an even number (from the definition of a even number, which is basically a number that can be divided by $2$). Now we know that in the number line, it goes odd, even, odd, even,..., and so if we add $1$ to this, then we get an odd number. We now have an odd number and an even number and so lets multiply them.

$$2k \cdot (2k + 1) = 2k(2k + 1) = 4k^2 + 2k) = 2(2k^2 + k)$$

We now have a random number $(2k^2 + k)$ multiplied by $2$ and so this is an even number (by definition of an even number).

My friend then said that we make the assumption that $2k \pm 1$ is odd because the number line alternates between odd and even numbers, how do know that this is true?

I was thinking how to prove this. Would it be by some form of induction?

EDIT: I think I need to change "number" to "integer" in this post don't I?

Answer

I wouldn't appeal to the fact that the integers alternate between even and odd.

I would say this:

An even number is defined as any integer of the form $2k$ for $k$ any integer (maybe excluding $k=0$).

An odd number is defined as any integer of the form $2k +1$ for any integer $k$.

You can just take this as the definition. And then (as you have) the proof that even times odd is even is just
$$2k(2k'+1) = 2[k(2k'+1)]$$
where of course $k(2k'+1)$ is just an integer.