I'll quickly go over my understanding of it:
If a number $n^2$ is even, then $n$ is even. The contrapositive is that if $n$ is not even, then $n^2$ is not even.
We represent n as $n=2p+1$. $n^2=4p^2 + 4p + 1 = 2(2p^2+2p) + 1$. We see that $n^2$ is odd. Therefore, the original statement must be true.
Now here's my question, set $n^2=2$. $n^2$ is even but $n=\sqrt2$ isn't. Or is it? I'm confused
On
Everything you wrote is correct up until the last line. If $n^2$ is even, then $$ n^2 = 2m \quad\text{ for some } m. $$
When you wrote $n^2 = 2$, you were (falsely) assuming that $m=1$.
By the way the even/odd language is only used when you're talking about integers (whole numbers), so there is no square root of $2$.
On
The original statement about "the square of a number" is true in a context where "number" means "integer." If that context is not clear, you had better say explicitly "the square of an integer". The square of any even integer can be written $2m$ where $m$ is an integer, but not every number that can be written as $2m$ where $m$ is an integer is a square of an integer of any kind.
Examples of squares of even integers written in the form $2m$:
$$\begin{align} 2^2 &= 4 = 2m & \text{where }\ & m=2. \\ 4^2 &= 16 = 2m & \text{where }\ & m=8. \\ 6^2 &= 36 = 2m & \text{where }\ & m=18. \\ 8^2 &= 64 = 2m & \text{where }\ & m=32. \\ \end{align}$$
Examples of $2m$ where $m$ is an integer, but $2m$ is not the square of any integer:
$$\begin{align} 2 = 2m & & \text{where }\ & m=1. & & \text{Not the square of any integer: }\ 1^2 < 2 < 2^2.\\ 6 = 2m & & \text{where }\ & m=3. & & \text{Not the square of any integer: }\ 2^2 < 6 < 3^2.\\ 8 = 2m & & \text{where }\ & m=4. & & \text{Not the square of any integer: }\ 2^2 < 8 < 3^2.\\ 10 = 2m & & \text{where }\ & m=5. & & \text{Not the square of any integer: }\ 3^2 < 10 < 4^2.\\ \end{align}$$
Both $n^2$ and $n$ must be integers for this theorem to hold.