Non isolated types of $\mathcal M$ cannot be isolated in $\mathcal N \succ \mathcal M$? Suppose $a \in M$ realizes a non-isolated type over $\emptyset$, and let $\mathcal N \succ \mathcal M$, furthermore let $|\mathcal M| = \aleph_o$ while $|\mathcal N| = \aleph_1$. Is it true that the type of $a$ also cannot be isolated in $\mathcal N$?
 A: Yes, this is true, assuming you're asking about whether or not the type of $a$ in $N$ over the empty set is isolated. For any theory $T$, the space of types over $\emptyset$ is defined without reference to any particular model. In particular the type (over $\emptyset$) of a tuple (or element) of a model either is or isn't isolated; it doesn't matter what model it came from. The last piece of the puzzle comes from the definition of elementary extension: if $N \succ M$, and $a \in M$, then a formula is true of $a$ in $M$ just if it's true in $N$. Therefore $a$ has the same type in both $M$ and $N$.
In general, if you're thinking about properties of types, it's best just to think of the type spaces themselves, without thinking about models other than as a place for parameters to live.
The business about cardinality is a red herring.
A: This is a detailed version of my comment.
Given a $\mathcal L$-structure $\mathfrak M$ and a subset $A$ of its base $M$, a complete $n$-type $p(\bar x)$ (with $\bar x$ a tuple of length $n$) is a maximal for inclusion set of $\mathcal L_A \cup\{\bar x\}$-sentences  such that $$ p(\bar x) \cup \operatorname{Th}(\mathfrak M, A) $$
is consistent (as a $\mathcal L_A \cup \{\bar x\}$-theory). The $n$th Stone space $S_n^{\mathfrak M}(A)$ is then the topological space of all those complete $n$-types together with the topology generated by the basic open sets
$$ \langle \varphi(\bar x) \rangle = \{p(\bar x) : \varphi(\bar x) \in p(\bar x) \} \qquad (\varphi(\bar x) \ \ \mathcal L_A\text{-formula}).$$

So given two $\mathcal L$-structures $\mathfrak M$ and $\mathfrak N$ containing both a common subset $A$, one has 
$$ S_n^{\mathfrak M}(A) \simeq S_n^{\mathfrak N}(A) \qquad (\text{as topological spaces})$$
as soon as $\operatorname{Th}(\mathfrak M,A) = \operatorname{Th}(\mathfrak N,A)$. This is notably the case when $A \subseteq \mathfrak M \preceq \mathfrak N$.
