@Zhen Lin's comment should really be promoted to an answer: he gives one good snappy argument. Here just for fun is another cute variant argument, depending on a Very Basic Theorem about effectively enumerable sets. [If you prefer, substitute 'recursively enumerable' throughout -- but the argument flies even deploying a more informal notion of effective enumerability.]
We are going to show, then, that
The set of truths of a sufficiently expressive language $L$ is not effectively enumerable.
Here a 'sufficiently expressive' language is one that can (i) express every effectively computable one-place numerical function, and (ii) can form wffs which quantify over numbers.
Proof Take the argument in stages. (i) The Very Basic Theorem about e.e. sets of numbers which I alluded to tells us
There is a set $K$ which is e.e. but whose complement $\overline{K}$ isn't.
Let's take that as proved. And now suppose the effectively computable function $k$ enumerates such a set $K$, so $n \in K$ iff $\exists x\,k(x) = n$, with the variable running over numbers.
(ii) Since $k$ is effectively computable, in any given sufficiently expressive arithmetical language $L$ there will be some wff of $L$ which expresses $k$: let's abbreviate that wff $\mathsf{K(x,y)}$. Then $k(m) = n$ just when $\mathsf{K(\overline{m},\overline{n})}$ is true.
(iii) By definition, a sufficiently expressive language can form wffs which quantify over numbers. So $\exists x\, k(x) = n$ just when $\mathsf{\exists x(\mathsf{Nat(x)} \land \mathsf{K}(x, \overline{n}))}$ is true (where $\mathsf{Nat(x)}$ stands in for whatever $L$-predicate might be needed to explicitly restrict $L$'s quantifiers to numbers).
(iv) So from (i) and (iii) we have $n \in K$ if and only if $\mathsf{\exists x(\mathsf{Nat(x)} \land
\mathsf{K}(x, \overline{n}))}$ is true; therefore, $n \in
\overline{K}$ if and only if $\mathsf{\neg\exists x(\mathsf{Nat(x)}
\land \mathsf{K}(x, \overline{n}))}$ is true.
(v) Now suppose for a moment that the set $\mathcal{T}$ of true sentences of $L$ is effectively enumerable. Then, given a description of the expression $\mathsf{K}$, we could run through the supposed effective enumeration of $\mathcal{T}$, and whenever we come across a truth of the type $\mathsf{\neg\exists x(\mathsf{Nat(x)} \land \mathsf{K}(x, \overline{n}))}$ for some $n$ -- and it will be effectively decidable if a wff has that particular syntactic form -- list the number $n$. That procedure would give us an effectively generated list of all the members of $\overline{K}$.
(vi) But by hypothesis $\overline{K}$ is not effectively enumerable. So $\mathcal{T}$ can't be effectively enumerable after all. Which is what we wanted to show.
