Formal language without grammar
I assume that by "grammar" you mean type-0 grammars. One can probably extend those to capture more languages.

Type-0 grammars are equivalent to Turing machines in expressive power. So, in order to show that a given language does not have a grammar, we can proceed as follows with suitable $L$:


  Assume there was a grammar for $L$. Then, we could semi-decide the word problem of $L$. That contradicts the known fact that $L$ is not recursively-enumerable.


Note the semi here; assuming a grammar does not give us decidability, so the halting problem is not a suitable candidate -- we need a problem/language that is not even semi-decidable/recursively enumerable. From your computability background you should know that the complement of the (special) halting language, namely

$\qquad \overline{H} = \{ \langle M \rangle \mid M \text{ loops on } \langle M \rangle \}$,

is not semi-decidable. Thus, by the reduction outlined above, there is no (Chomsky) grammar for $\overline{H}$.