Are the implications of the diagonalization language different from those of the halting problem?
I think that "limits of computation" is a good entry point. Usually, computer scientists learn how to solve problems in an logical and algorithmic way. The whole concept of undecidability is to show that there are problems which cannot be solved in this way, which may sounds surprising but also exciting, that there is an actual limit to what is possible with computers (even with infinite memory).

You actually do not need the diagonalization language to show that there are undecidable problems as this follows already from a combinatorical argument:
You can enumerate the set of all Turing machines (sometimes called Gödelization). Thus, you have only countable many decidable languages. But the set $2^{\{0, 1\}^\ast}$, i.e. the set of all languages over $\{0, 1\}$, is uncountable, so there have to be undecidable languages.

The diagonalization language (I assume $D = \{\langle M \rangle \mid M \text{ halts on } \langle M \rangle\}$ is by all means very constructed but the proof that this language is undecidable is a good stating point because the proof via diagonalization is easy and the usual variations of the halting problems are not really far away. The next steps are probably Turing reductions from this language to some of these variations.