Proving the completeness of a theory $\Gamma$ Given a set of sentences $\Gamma$ in a first-order-language $\mathcal{L}$, such that for all structures $\mathcal{A}=(A,\ldots)$ and $\mathcal{B}=(B,\ldots)$, if both $\mathcal{A}$ and ${\cal B}$ realize $\Gamma$, and $A$ and $B$ are at most countable sets, then $\cal A\cong \cal B$.
I have to prove that $\Gamma$ is complete.
Should I now assume that, for a given $\cal L$-sentence $\varphi$, not $\Gamma\models\varphi$, and derive that $\Gamma\models \neg\varphi$ must hold? I have tried this, but I don't have a clue how to do it.
I've also tried considering two random structures $\cal A,\cal B$ that realize $\Gamma$, but from there I can't derive anything.
Is one of these approaches right, or do I have to try something else?
 A: Hint $\Gamma\models\varphi$ iff $\mathcal{A}\models\varphi$ for all $\mathcal{A}$ which realizes $\Gamma$. $\Gamma$ has just one (at most) countable model. If $\Gamma$ has infinite model and if $\mathcal{A}$ is an uncountable model of $\Gamma$, you can find a countable one which is elementary equivalent to $\mathcal{A}$.

If $\mathcal{L}$ can be uncountable, we can construct a counterexample of the problem.
Consider the language consists of constant symbols $a$, $b$ and $c_r$ for each real number $r$. Let $\Gamma$ be the set of formulas which consists of the formulas
$$a=b\leftrightarrow c_r=c_s \qquad \text{for each }r\neq s$$
and
$$a=b\leftrightarrow \forall x\forall y (x=y).$$
If $\mathcal{A}\models \Gamma$ and $a^\mathcal{A}=b^\mathcal{A}$, then $\mathcal{A}$ has just single element, and if $\mathcal{A}\models \Gamma$ and $a^\mathcal{A}\neq b^\mathcal{A}$ then $\mathcal{A}$ must be uncountable. So every model of $\Gamma$ which is at most countable is isomorphic, but $\Gamma$ is not complete since neither $a=b$ nor $a\neq b$ is derived from $\Gamma$. 
A: Suppose $\Gamma$ is not complete. Then there is a $\varphi$ such that $\Gamma,\varphi$ and $\Gamma,\neg\varphi$ are both consistent. By Gödel both have models, and then by the downward Löwenheim-Skolem theorem each has an at most countable model. These models are in particular models of $\Gamma$ ...
