Non standard analysis relies on using sets that have the same properties as sets of real numbers to transfer many good theorems over. Every set $S\in P(\Bbb R)$ has a unique extension $S^* \in P(\Bbb R^*)$. $P(\Bbb R)^*$ clearly has the same number of elements as $P(\Bbb R)$ seeing as it’s defined by a bijection, but set of internal sets is a superset of that and a subset of $P(\Bbb R^*)$.
$|P(\Bbb R)| < |I|< |P(\Bbb R^*)|$ where I is the set of internal subsets. This sandwiched inequality along with the fact that the leftmost and rightmost sides are equal, should prove that $I$ has cardinality $2^c$.
Is there any error in this proof or something I missed that could make it faster?