# Proving the diagonalization of $U \subset \mathbb{N} \times \mathbb{N}$ that is an universal set for all enumerable sets of naturals is undecidable

Let $$U \subset \mathbb{N} \times \mathbb{N}$$, and let $$U_n = \{ x \mid (n, x) \in U \}$$. Let's call $$U$$ universal for some class $$\mathcal{S}$$ of subsets of $$\mathbb{N}$$ if $$S \in \mathcal{S} \iff \exists n : S = U_n$$. In other words, the set of "cuts" of $$U$$ contains all the sets in $$\mathcal{S}$$ and nothing more.

Now, let $$U$$ be universal for the class of all enumerable subsets of $$\mathbb{N}$$. Prove that $$K = \{ x \mid (x, x) \in U \}$$ is enumerable but not decidable.

So, my proof sketch is as follows:

Enumerability is obvious, so let's focus on decidability. I know that there exists an enumerable undecidable $$F \subset \mathbb{N}$$. By definition of $$U$$, it must be contained in a cut with some index $$k$$. Now, deciding whether $$(k, k) \in U$$ is analogous to deciding whether $$k \in F$$. Any function $$f$$ for the latter is not computable, hence, the whole function $$u$$ that decides whether $$(x, x) \in U$$ for arbitrary $$x$$ is not computable, so $$K$$ is undecidable.

Firstly, the proposition that $$f$$ being undecidable implies $$u$$ is undecidable seems to follow from the definitions in my book (where a set is decidable if its characteristic function is computable), but, intuitively, it just feels unjustified.
Secondly, have I proved too much? A similar argument could be applied to show that any $$K' = \{ x \mid (x, f(x)) \in U \}$$ is undecidable for arbitrary $$f$$ — it's sufficient for $$x$$ to range over the whole $$U$$ to capture the undecidable set that's contained in its cut. Is that correct?
Indeed, the usual diagonalization approach just works: assume $$K$$ is decidable, then $$\overline{K} = \{ x \mid (x, x) \not \in U \}$$ is also decidable, then $$\overline{K}$$ is enumerable, then it's contained, let's say, in $$U_x$$ (by definition of $$U$$). Then the contradiction easily follows.