I assume you know the Ryll-Nardzewski theorem, at least in the form which states that for a countable and complete theory $T$ with infinite models, $T$ is $\aleph_0$-categorical if and only if $S_n(T)$ is finite for all $n$.
What we know is that for all $n$, there is a model $M$ which only realizes finitely many types in $S_n(T)$.
Let's fix an arbitrary $n$ and show that $S_n(T)$ is finite. Let $M$ be a model of $T$ which only realizes finitely many complete types in $S_n(T)$, say $p_1,\dots,p_k$. For each $1\leq i \leq k$, let $\varphi_i$ be a formula in $p_i$ such that $\lnot \varphi_i\in p_j$ for all $j\neq i$. You can find $\varphi_i$ by picking, for each $j\neq i$, some formula $\psi_j\in p_i$ such that $\lnot \psi_j\in p_j$, and defining $\varphi_i = \bigwedge_{j\neq i} \psi_j$.
Now every $n$-tuple from $M$ realizes one of the types $p_i$, so $$M\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$
And for each $i$, if an $n$-tuple from $M$ satisfies $\varphi_i$, then the tuple must realize $p_i$, so for all $\theta\in p_i$, $$M\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$
Now $T$ is complete, so any sentence satisfied in any model is actually satisfied in every model. So we have
$$T\models \forall x_1\dots\forall x_n\, \bigvee_{i=1}^k \varphi_i.$$
And for each $i$ and each $\theta\in p_i$, $$T\models \forall x_1\dots\forall x_n\, (\varphi_i\rightarrow \theta).$$
This shows that every model of $T$ only realizes the types $p_1,\dots,p_k$, since any $n$-tuple from any model of $T$ must satisfy $\varphi_i$ for some $1\leq i \leq k$, and thus must realize $p_i$. So in fact $S_n(T) = \{p_1,\dots,p_k\}$ is finite.
Since $n$ was arbitrary, $S_n(T)$ is finite for all $n$, so $T$ is $\aleph_0$-categorical.
