Confirming a easy proof: the product of two consecutive numbers is always even. Can someone confirm if my prove is right?
Theorem. The product of two consecutive integers is always even.
Proof. Define a number $n$ such that $n:=2k$ where $k\in\mathbb{Z}$, this ensures that $n$ is an even number. Define a second number $p$ such that $p:=n+1=2k+1$ this ensures that $p$ is an odd number and the numbers $n$ and $p$ are consecutive because they differ by $1$. There product is given by:
$$np=2k(2k+1)=2(2k^2+k)=4k^2+2k\tag1$$
Which is clearly a multiple of $2$.
Define a number $h$ such that $h:=2m+1$ where $m\in\mathbb{Z}$, this ensures that $h$ is an odd number. Define a second number $z$ such that $z:=n-1=2m$ this ensures that $z$ is an even number and the numbers $h$ and $z$ are consecutive because they differ by $1$. There product is given by:
$$hz=2m(2m+1)=2(2m^2+m)=4m^2+2m\tag2$$
Which is clearly a multiple of $2$, this proves that $hz$ is an even number.
This proves this theorem.

If I am wrong how can I make my proof valid?


Style and sufficiency of proof generally depends on where you take the "floor of certainty" to be - what facts and statements can be taken as true and what needs to be demonstrated.
With a simple theorem like this you might say that we should set the floor a little lower, that is provide an exposition of even simple steps in the proof. For example, the fact that alternating integers are odd and even might need to be supported - or not.
Commenting on your proof, then, in the knowledge that this is a little subjective:

*

*Breaking into the cases of lower number even and lower number odd looks good, and you could do this more clearly.

*You don't need to define a separate number for $n{+}1$. It doesn't add clarity and could easily confuse the matter.

*You can continue to use $n$ and $n{+}1$ as the successive numbers in the two cases - again limit your new defined numbers, for example to where they will interact with each other; there's no such interaction between your cases.

So using your method I would write:

Theorem. The product of two consecutive integers is always even.
Proof. Consider successive integers $n$ and $n{+}1$ with product $n(n{+}1)$ in two cases, (1) where $n$ is even and (2) where $n$ is odd.
Case 1, $n$ even:
Find $k\in\mathbb{Z}$ such that $n=2k$. Now the desired product is given by:
$$n(n+1)=2k(2k+1)=2(k(2k+1))\tag1$$
And since $k(2k+1)\in\mathbb{Z}$ this is thus a multiple of $2$ and even as required.
Case 2, $n$ odd:
Find $k\in\mathbb{Z}$ such that $n=2k+1$. Now the desired product is given by:
$$n(n+1)=(2k+1)(2k+2)=2(2k+1)(k+1)\tag 2$$
And since $(2k+1)(k+1)\in\mathbb{Z}$ the product is again a multiple of $2$ and even as required.
Since both cases demonstrate an even product, this proves this theorem.

