Logic question linking $\omega$-categoricalness to completeness Please check my attempted "answer" below. Any corrections are gladly welcomed!
I am stuck with this problem:
Assume that $T$ is a consistent theory of a first order language $\mathcal{L}_\mathcal{A}$. Assume that every model of $T$ is infinite. Show that if $T$ is $\omega$-categorical, then $T$ is complete.

The problem I am facing is that, according to my lecture notes, the definition of $\omega$-categorical requires $T$ to be complete in the first place.
 (Definition: Suppose that $T$ is a complete, consistent theory. $T$ is $\omega$-categorical if and only if any two countable models of $T$ are isomorphic.)

Sincere thanks for any help!
 A: Better: A consistent theory $T$ is $\omega$-categorical if and only if any two countable models of T are isomorphic.
To show if $T$ is $\omega$-categorical then complete, prove the contrapositive. Suppose $T$ is not complete. Then for some $\varphi$, $T + \varphi$ and $T + \neg\varphi$ are both consistent, so by Löwenheim-Skolem both have countable models which can't be isomorphic ...
A: Thanks Peter Smith. I attempt to elaborate more on his answer (to improve my understanding and to practice logic notation, which I am weak and unfamiliar with). Please give any comments on whether it is correct, it is the only way that I can check if I am on the right track. Sincere thanks!

Suppose to the contrary $T$ is not complete.
Then there exists a sentence $\varphi$ such that $\varphi\notin T$ and $(\neg\varphi)\notin T$.
Then, we claim that $T\cup\{\varphi\}$ and $T\cup\{\neg\varphi\}$ are both consistent. (not sure how to show that)
Hence, by Gödel's completeness theorem, $T\cup\{\varphi\}$ and $T\cup\{\neg\varphi\}$ are both satisfiable.
There exists $\mathcal{M}_1$, $\mathcal{M}_2$ such that $\mathcal{M}_1\models T\cup\{\varphi\}$ and $\mathcal{M}_2\models T\cup\{\neg\varphi\}$.
By Lowenheim-Skolem, there exists elementary substructures $\mathcal{N}_1\leq\mathcal{M}_1$ and $\mathcal{N}_2\leq\mathcal{M}_2$, with $\mathcal{N}_1,\mathcal{N}_2$ countable.
Then, $\mathcal{N}_1\models T\cup\{\varphi\}$ and $\mathcal{N}_2\models T\cup\{\neg\varphi\}$
In particular, we have $\mathcal{N}_1\models T$ and $\mathcal{N}_2\models T$, 
so $\mathcal{N}_1$ and $\mathcal{N}_2$ are two countable models.
But $\mathcal{N}_1\models\varphi$ while $\mathcal{N}_2\models (\neg\varphi)$.
Hence, $\mathcal{N}_1$ and $\mathcal{N}_2$ are not isomorphic. (Contradiction)
Hence $T$ is complete.
