Is the set of Gödel numbers of computable constant functions recursively enumerable?
Your intuition is good: checking whether $x \in S$ is stronger than checking $x \in K$ ($K$ the halting problem, i.e. $K = \{ x \mid f_x(x) \text{ halts}\}$). In other words,

$\qquad\displaystyle \langle \operatorname{sgn} \circ f\rangle \in S \implies f \in K$.

However, the reverse does not hold and no similar implication works for $\overline{K}$, the reduction partner we really want (as it's not semi-decidable).



A small interlude: the title you chose does not actually fit the question! The set of all constant functions is in fact recursively enumerable, e.g. by

$\qquad\displaystyle \varphi_i(x) = i$

which is clearly a computable function. Many sets of functions are enumerable like this, but not all.

What you are really asking is: given a Gödel numbering (e.g. an encoding of all Turing machines), what about the set of all indices (read: programs) that compute this here set of functions? That's another thing entirely because of the properties such a Gödel numbering has. The distinction is important, see e.g. here.



The basic idea for a reduction is always this: build a function that depends on $x$ and whose encoding is in $S$ if and only if $x \in \overline{K}$ -- then we got a deal.

So, consider

$\qquad\displaystyle g_x(n) = 
    \begin{cases}
        n, &amp;f_x(x) \text{ halts after at most $n$ steps } \\
        1, &amp;\text{else}
    \end{cases}
$

which is clearly computable. Note furthermore that given $x$, we can compute $y$ with $f_y = g_x$; such compilation is possible because we have (or can assume) a Gödel numbering. Now, clearly

$\qquad \langle g_x \rangle \in S \iff x \in \overline{K}$

holds; thus $S$ can't be semi-decidable since $\overline{K}$ would then be semi-decidable, too, contradicting what we know.