This is a naive, kind of informal argument.
Suppose we have a language with just one predicate $P$ and constants $a_{1}, a_{2}, a_{3}$ and so on. Suppose also that we have countably many worlds $1, 2, 3$ and so on. Without loss of generality, we assume that $Pa_{1}$ holds in all the worlds. That is, the set of worlds in which $Pa_{1}$ holds is $N$.
Now, $Pa_{1}\&Pa_{2}$ will hold in fewer or at most the same set of worlds. For purpose of argument ,we take that it holds in fewer worlds. So the Set $S_{2}$ of worlds in which $Pa_{1}\&Pa_{2}$ holds is a strict subset of $N$. We continue this process, assuming that strict inclusion holds at every stage.
So we have a nested series of subsets $S_{i}$ of $N$, such that $Pa_{1}\&Pa_{2}\&\dotsb\&Pa_{i}$ holds in $S_{i}$.
Clearly, the set of worlds in which the universal sentence ($\forall nPn$?) AnPn holds Is the intersection of all the nested sets $S_{i}$ which is the null set. That is , there is no world in which the universal sentence is true.
Alternately, we have to give up the assumption of strict inclusion of The successive sets $S_i$, so that
There is some $k$, such that for all $i$ greater than $k$, $S_i =S_k$, which means
That $Pa_i$ for $i > k$ is true in all worlds in which $Pa_1\&Pa_2\&\dotsb\&P_k$ is true.
Now my question is ( assuming improbably that there is no flaw in the argument above)
Why cannot this argument go through formally? Is it something to do with the Lowenheim-Skolem theorem ?
There is a flaw in the argument. Consider the following sets $S_1,S_2, \dots$.
$S_1$ is all of $\mathbb{N}$.
$S_2$ is all of $\mathbb{N}$ except $1$.
$S_3$ is all of $\mathbb{N}$ except $1$ and $3$.
$S_4$ is all of $\mathbb{N}$ except $1$, $3$, and $5$.
And so on.
We have strict inclusion at all stages.
The intersection of the $S_i$ is the set of even numbers. It is very much non-empty.