There exist countably many enumerable disjoint and inseparable sets
Let $\phi$ be an enumeration of the set of recursive partial functions.

For each $i\in\mathbb N$, let $X_i = \{ n\ |\ \phi_n(0)=i \}$.

These sets are pairwise disjoint by construction. Each $X_i$ is recursively enumerable but not recursive.

These sets are pairwise inseparable. By contradiction assume there exist $i\neq j$, and a recursive set $A$ with $X_i \subseteq A$ and $X_j \subseteq \mathbb N\setminus A$. We reach a contradiction as follows. The function

$$
g(k,y) = \begin{cases}
  j &amp; \mbox{if } k \in A \\
  i &amp; \mbox{o.w.}
\end{cases}
$$
is computable, hence $g=\phi_c$ for some $c$. Now 
exploit the s-m-n lemma to define $h(k) = \mathsf s(c,k)$, so that
$\phi_{h(k)}(y) = g(k,y)$.
Since $h$ is recursive total, by the second recursion theorem for
some $k$ we have $\phi_{h(k)} = \phi_k$. 
Let us take one such $k$.
Finally,

$$
\phi_k(0) = \phi_{h(k)}(0) = g(k,0) = \begin{cases}
  j &amp; \mbox{if } k \in A \\
  i &amp; \mbox{o.w.}
\end{cases}
$$

Now, the result above is either $j$ or $i$. If it is $j$, then $k\in X_j$, hence $k\notin A$, hence the result above is $i$ -- contradiction.
If it is $i$, then $k\in X_i$, hence $k\in A$, hence the result above is $j$ -- contradiction.