How to prove that not all recursive sets are in $\mathsf{P}$?

Question

I am interested in finding simple proof (without time hierarchy theorem) of the following fact:

To show that there is a recursive set that is not in class $\mathsf{P}$.

Answer 1

The class $\mathsf P$ is effectively enumerable: If we define $$ A_k = \{n\mid \text{Turing machine number $k$ accepts $n$ in at most $n^k+k$ steps}\}$$ then every set in $\mathsf P$ is $A_k$ for some $k$.

This works because every real polynomial is pointwise smaller than something of the form $x^k+k$, and every Turing machine has (in effect) arbitrarily large indices -- namely, we can add unreachable states to it until the index is as large as we want it. (We could avoid this reasoning at the cost of introducing a tupling function and corresponding projections).

Now we can diagonalize to find $$ D = \{ n \mid \text{Turing machine number $n$ does not accept $n$ in at most $n^n+n$ steps}\} $$ which is clearly recursive, but by construction differs from all the $A_k$s and so is not in $\mathsf P$.

Whether this proof is significantly different from just applying the time hierarchy theorem is, I suppose, in the eye of the beholder.