It is definitely way too easy to accidentally confuse "x is true" and "x is proven to be true" on topics of undecidability, and this is one of those times where the answer is in fact obvious, but it takes you 15 minutes to realize that it's obvious.

The halting problem is uncomputable (by diagonalization)--that is no computable function can solve the halting problem, rather than the weaker statement that we cannot assert that "this" function solves it. As a result, if somebody asserts a function that purports to solve the halting problem, then we can definitively say that either it isn't computable, or the thing it solves isn't the halting problem.

For the Busy Beaver function, it's obvious that the resulting program (if it existed) would solve the halting problem, so clearly it's not a computable function, and analyzing what part of it isn't computable leads you to the Busy Beaver as the only option.

For the D(N) function... well, since we assume D(N) ≥ BB(N), D(N) is still an upper bound on the number of steps a halting TM could run, so the resulting program using D(N) in lieu of BB(N) would still solve the halting problem, which forces us to conclude that it's not computable.

A different argument that may make more sense is this:

Consider the program "if machine M has not halted after f(sizeof M) steps, print not halt, else print halt." If f is a computable function, then the program is clearly computable. But since no computable program can solve the halting problem, we know that this program cannot either. Therefore, for every computable function f, there must exist some machine M such that M halts only after more than f(sizeof M) steps. In other words, f cannot be an upper bound on the Busy Beaver function.