Suppose $T$ is a $\aleph_0$ stable theory in a countable language. Fix some set $A$.
Let $S = \{ R[p]:p \in S(A), p$ is non-algebraic$\}$ Let $\alpha = min(S)$ and let $p \in S$ be such that $R[p] = \alpha$. Then show that for any $B$ such that $A \subseteq B$, and $q_1, q_2 \in S(B)$ non-algebraic such that $p_0 \subseteq q_1, q_2$, then we have $q_1 = q_2$.
Definition of R
$R$ here is a modified binary version of Morley rank. Given a type $p$, and an ordinal $\alpha$, $R[p] \geq \alpha + 1$ iff for every finite $p_0 \subseteq p$ there is $\varphi$ such that $R[p_0 \cup \{\varphi\}], R[p_0 \cup \{\neg \varphi\}] \geq \alpha$.
The base case and limit case are defined as you'd expect (base case: $\geq 0$ if consistent/realized, and limit case: $\geq \alpha$ iff $\geq \beta$ for every $\beta < \alpha$). I also supressed information relating to parameters and such to keep it from getting too wordy. Let me know if you'd like me to add details.
A few basic results that are basically the same as in Morley Rank:
$p \vdash q \implies R[p] \leq R[q]$
$R$ has finite character (i.e every type has a finite subtype with the same rank)
$R$ is invariant under automorphisms
Uniqueness: if $R[p] < \infty$, $p \subseteq q_1, q_2$ where $q_1, q_2$ are complete types over bigger domains,and $R[p] = R[q_1] = R[q_2]$, then $q_1 = q_2$.
Stability: $T$ is $\aleph_0$ stable iff $R[x = x] < \omega_1$ iff $R[x = x] < \infty$.
As you can see it's quite similar to Morley rank.
I've been working on it for a bit with no avail.
You're probably having trouble proving this because it is not true as stated. Here's a counterexample.
Let $T$ be the theory of an equivalence relation $E$ with two infinite classes. Pick representatives $b_1$ and $b_2$ for the two classes. Let $A=\varnothing$, and let $B=\{b_1,b_2\}$.
Now there is a unique $1$-type $p$ in $S(A)$, which is of course the non-algebraic type of minimal rank (I think its rank is $\omega+1$), but $p$ has two distinct non-algebraic extensions to $S(B)$, isolated by $(xEb_1\wedge x\neq b_1)$ and $(xEb_2\wedge x\neq b_2)$ respectively.