I'm trying to prove the following theorem,
Let $(X, \leq)$ be a complete lattice. If $\phi:X \to X$ is order preserving, it has a fixed point.
So far, I've done the following:
I've defined the set $A := \{a \in X : a \leq \phi(a)\} $. Since $\inf{X} \in A$, $A$ is non empty and therefore we can consider $x_0 := \sup{A}$. Now, since $\phi$ is order preserving,
$$ a \leq \phi(a) \leq \phi(x_0) \ \ \ (\forall a \in A) $$
This shows that $\phi(x_0)$ is an upper bound of $A$, and therefore $x_0 \leq \phi(x_0)$. Again, because $\phi$ is order preserving, this implies $\phi(x_0) \leq \phi(\phi(x_0))$ which means that $\phi(x_0) \in A$, so $\phi(x_0) \leq x_0$, (supposedly) completing the proof.
Is this correct?
Yes, this is quite correct. Why do you doubt this? All steps have been justified, IMHO. The last step is just $\phi(x_0) \le x_0$ and $x_0 \le \phi(x_0)$ implies $\phi(x_0) = x_0$ by the axioms for partial orders, so $x_0$ is a fixpoint for $\phi$. (Why the supposedly?)