First-order logic - Does there exist a sentence that is satisfiable by all infinite models and only by them?

Question

Prove or Disprove:
There is $\boldsymbol{no}$ alphabet $\Sigma$ and closed formula (no free variables) $\varphi$ above $\Sigma$, such that for any Model $M$ it holds that $M\models\varphi\iff\,|D^M|=\infty$.

I'm not sure what is the classic way to approach this. I do think that the statement is correct. I tried the next proof:
Assume by contradiction that there exist $\Sigma,\,\varphi$ as mention above. So it also holds that for any model $M$:
$M\models\neg\varphi\iff\,|D^M|<\infty$
Let $M$ be a model for $\Sigma$ such that $D^M=\{1,...,n\}$ and $M$ defines functions and relations in $\Sigma$ arbitrary.
$M$ domain is of finite size so we have $M\models \neg\varphi$
Define $M'$ to be an expansion of $M$:$\,\,$ $D^{M'}=\mathbb{N}$, and for every $k\in\mathbb{N}$,$\,\,\,M'$ treats $k$ the same way $M$ treats $k\,\,mod\,\,n$.
Then I proved by induction on $\varphi$'s structure that:
$M\models\neg\varphi$ $\iff$ $M'\models\neg\varphi$
and becuase we know that $M\models\neg\varphi$ we have $M'\models\neg\varphi$ which is contradiction, becuase $M'$ domain is of infinite size.

Is my proof correct? Or should I build additional model $N$, of finite domain, isomorphic to $M'$, and show with $N$ the contradiction?
Is there a better known way to prove this? Sometimes there are claims that there is a classic way to prove / disprove, I wonder if in this case as well.

Would appreciate help! :-)

Answer 1

Probably the easiest way to prove this is using the compactness theorem: assume for contradiction that $\varphi$ is such a sentence. For every $n\in\mathbb N$, let $E_n$ denote the sentence $\exists x_1,\dots,x_n\bigwedge_{i<j}x_i\ne x_j$, which asserts that there are at least $n$ elements. Since every finite subset of $S=\{\neg \varphi\}\cup\{E_n:n \in \mathbb N\}$ is true in a sufficiently large finite structure, $S$ is consistent by the compactness theorem. But a model $M\models S$ is an infinite structure that does not satisfy $\varphi$, a contradiction.