Let $\alpha$ be a composite natural number not equal to 4. Show that $\exists m,n \in \mathbb{N}$ such that $ 1 < m < n < \alpha$ and $\alpha|mn$. This is my proof so far. Split it up into cases, one where $\alpha$ is the square of a prime and one where it is not.
Case 1) If $\alpha = p^2$ is the square of a prime. Then naturally $p < \alpha$ and we can choose a number $1 < k < p$ such that $kp < \alpha$. Hence $p^2|kp^2$, so $\alpha|mn$ where $m=p,n=kp$
Case 2) If $\alpha$ is not the square of a prime. Let X be the set of all prime factors of $\alpha$. By the well ordering principle, there exists a smallest element in X, lets name it $\beta$. Since $\beta|\alpha$ there exists a $\gamma$ such that $\beta\gamma = \alpha$ which means that $\alpha|mn$ where $m = \beta,n=\gamma$.
I was wondering if this proof was correct, and if it is sufficient? Did I make any errors?
Your proof is almost perfect. You should just note in case 2 that $\gamma \neq \beta$. The fact that $\beta, \gamma < \alpha$ is implied by $\beta\gamma = \alpha$, but you could point it out if you're feeling pedantic.