Incomputable sets of low degree vs Rices theorem?
This is not an answer, but way too long to be a comment.  I just wanted to cite Robert I. Soare's Recursively Enumerable Sets and Degrees, Chapter V: Simple Sets and Post's Problem.

1: Immune Sets, Simple Sets, and Post's Construction
The only r.e. degrees constructed so far are $\textbf{0}$ and
$\textbf{0}'$.  Post's Problem was to construct other r.e. degrees,
i.e., to construct an r.e. set $A$ such that
$\emptyset \lt_\text{T} A \lt_\text{T} \emptyset'$.  Post's Program
for constructing such a set $A$ was to ﬁnd some easily deﬁnable property
on $\overline{A}$ (compatible with $A$ being nonrecursive) which
guarantees incompleteness ($K \not\leq_\text{T} A$).
Recall (from Theorem II.4.5) that $K$ (and hence any creative set) has
an inﬁnite r.e. set (and therefore inﬁnitely many inﬁnite r.e. sets)
contained in its complement. Post's idea for constructing $A$ incomplete
was to make $\overline{A}$ sufﬁciently &quot;thin&quot; with respect to
containment of inﬁnite r.e. sets so that $\overline{K}$ could not be
Turing reduced to $\overline{A}$.
1.1 Deﬁnition. (i) A set is immune if it is infinite but contains no
infinite r.e. set.
(ii) A set $A$ is simple if $A$ is r.e. and $\overline{A}$ is immune.
1.2 Proposition. If $A$ is simple then:
(i) $A$ is not recursive;
(ii) $A$ is not creative;
(iii) $A$ is not $m$-complete (i.e., $K \not\leq_m A$).
Proof. (i) $\overline{A}$ cannot be r.e., since otherwise it contains an
inﬁnite r.e. set (namely $\overline{A}$).
(ii) The complement of a creative set does contain an inﬁnite r.e. set (by
Theorem II.4.5).
(iii) If $K \leq_m A$, then $A$ is creative (by Corollary II.4.8). $\Box$
1.3 Theorem (Post [1944]). There exists a simple set $S$.
Proof. Let $A = \{(e,x) \;:\; x \in W_e \;\&amp;\; x &gt; 2e \}$.  Now $A$ is
$\Sigma_1$ and hence r.e. Let $\psi$ be any p.r. selector function for
$A$ (in the sense of Theorem II.1.6) and let $S = \text{rng}\; \psi$.
(Intuitively, enumerate $W_e$ until the ﬁrst element $\psi(e) &gt; 2e$
appears in $W_e$, and then put $\psi(e)$ into $S$.) The following facts
give the simplicity of $S$.
(1) $S$ is r.e. ($S$ is the range or a p.r. function.)
(2) $\overline{S}$ is inﬁnite.  To see this, note that $S$ contains at
most $e$ elements out of $\{0,1,2,\dots,2e\}$, namely $\psi(0),\psi(1), \dots, \psi(e-1)$. Hence $e + 1 \leq \text{card}(\overline{S} \upharpoonright (2e+1))$, so $\overline{S}$ is inﬁnite.
(3) If $W_e$, is inﬁnite, then $W_e \cap S \neq \emptyset$, because
$(e,x) \in A$ for some $x &gt; 2e$ and so $\psi(e)$ is deﬁned and $\psi(e) \in S \cap W_e$. $\Box$