TITLE: Is this a good way to explicate Skolem's Paradox?
QUESTION [9 upvotes]: Skolems Paradox shows an ostensible conflict between Cantor's Thoerem (CT) and the downward Löwenheim–Skolem Theorem (ST).
CT: for any set $A$, the powerset of $A$, $P(A)$, has a strictly greater cardinality than $A$.  Cardinality is in terms of bijections: two sets have the same cardinality iff there exists a bijection between them.  A set $A$ is countable iff there exists a objection between $A$ and the set of naturals $\omega$.  Since (by CT) no function surjects $\omega$ onto its powerset, we learn that $P(\omega)$ is uncountable.  Thus CT generally tells us some sets are uncountable.
ST: If a countable first-order theory has an infinite model, it has a countable model.  The standard axiomatization of set theory, ZFC, is such a theory.
Assume ZFC has a model (which must be infinite).  
By ST, ZFC has a countable model .  
By CT, we can deduce the existence of uncountable sets from ZFC.  
Therefore, there must be some set A ∈  such that  satisfies $A$ is uncountable.  That is, there is no bijection $f \in$  between $A$ and $\omega$.
However, since  has a countable domain, there are only countably many elements available to be members of A.
Thus A appears both countable and uncountable.

Some good links I found:
1. How did first-order logic come to be the dominant formal logic? (and comments)

REPLY [12 votes]: Indeed $\mathfrak M$ sees only a small fraction of the universe, in particular he does not see the bijection between $\omega$ and itself, or between $\omega$ and some of the members of $\mathfrak M$.
However it is perfectly fine to have a model which is countable and has elements which are uncountable.
If you start by taking the real numbers and using it to generate a countable model, you will end up with a countable model of ZFC (of course it would not know of its own countability) in which there is a really uncountable set.
Now recall that $\mathfrak M\models A\text{ is countable}$ if and only if there is $f\in\mathfrak M$ which is a bijection between $A$ and $\omega$. If $\mathfrak M$ does not know about such $f$, $\mathfrak M\models A\text{ is uncountable}$. This is the heart of internal-external points of view.
If, on the other hand, we require $\mathfrak M$ to be transitive then every element of $\mathfrak M$ has to be countable as well (since it is a subset of $\mathfrak M$ by transitivity).