Show that doesn't exists $\Delta$ such that it is satisfable iff assigment has finite number of $1's$. It means that we assign only finitely many $1's$ for variables and from this fact we would like to conclude that $\Delta$ is satisfable. And if $\Delta$ is satisfable then assigment constains only finite unmber of $1's$. Number of variables is infinite: $a_1,a_2,...$. We are in predicate-logic (no quantifiers).
My idea is following:
Let's suppose that such $\Delta$ exists. Lets consider any finite subset of $\Delta_0\subseteq_{\text{fin}}\Delta$ such that it may be contradictory - finite subset with this property must exists - if it doesn't exists $\Delta$ is satisfable regardless to valuation of variables - then we have contradiction and show that such $\Delta$ doesn't exists.
So, we have finite $\Delta_0$ such that it may be contradictory - we force contradictory - we use only finitely many values $1$ because subset is finite. Rest of variables we set to $0$. However, $\Delta$ is unsatisfable becuase there exists unsataisfable finite susbet (compacntess theorem). Although, we assign only finite number of ones, $\Delta$ is unsatisfable.
Am I ok ?
From the comment thread, I think a correct solution (or rather, a clear-ish statement of the idea in your post, which is correct) has emerged; but let me state it cleanly. In particular, we can avoid the Compactness Theorem entirely.
You argue as follows:
Suppose $\Delta$ characterizes the assignments with finitely many $1$s. Then consider the assignment $\nu$ which sets all variables to $1$; this assignment has infinitely many $1$s, so we can find some element $\varphi$ of $\Delta$ not satisfied by $\nu$.
Alright, so look at this $\varphi$. Any assignments which agree on the variables in $\varphi$, agree on whether they satisfy $\varphi$ or not. Since $\varphi$ is a single sentence, it uses only finitely many variables; so, letting $\mu$ be the assignment which agrees with $\nu$ on those variables (i.e. sets them to $1$) and sets all other variables to $0$, since $\nu$ doesn't satisfy $\varphi$ we must also have that $\mu$ doesn't satisfy $\varphi$.
But then $\mu$ doesn't satisfy $\Delta$, since $\varphi\in \Delta$; a contradiction, since $\mu$ has only finitely many $1$s.
Note that this argument did not use compactness at all. This was one of the points that confused me in the comment thread - I didn't understand how compactness was entering into the proof.
There is also an argument using compactness; I suspect this is the intended solution.
Given such a $\Delta$, consider the theory $$\Delta\cup\{a_i: i\in \mathbb{N}\}.$$ This is finitely satisfiable (why?), hence has a model; but that model has infinitely many $1$s, contradicting the assumption on $\Delta$.