Is an infinite language of halting TM is in $RE$? in $RE \setminus R$?
The empty set $\emptyset$ satisfies trivially the constraint as all the TM it
contains (actually none) halt for all input. But the emptys set is
also in $R$ since it is very easy to decide whether a TM is in it: the
answer is always NO. But unfortunately, this example cannot be used, because we hav a requirement that the set $L$ of TM be infinite (Thanks to @FrançoisG for reminding me).

So we choose another simple counter example, the set $L_{\emptyset,0}$ of TM recognizing the empty language that halt and reject after 0 steps, but may have a lot of useless states and transitions. There is also an infinite number of such machines, and it is easy to see by inspection of its transitions whether a TM is in $L_{\emptyset,0}$. Thus $L_{\emptyset,0}$ is infinite and recursive.

Since the set $L_{\emptyset,0}$ is recursive, there is a computable bijection   $\phi: \mathbb N\to L_{\emptyset,0}$

Then given any subset $S$ of $\mathbb N$, we can get get with the bijection $\phi$ a subset $\phi(S)\subset L_{\emptyset,0}$ which has the same level of computability.

Thus if we take any non-recursive RE subset $S_{RE\setminus R}\in\mathbb N$, we have $\phi(S_{RE\setminus R})\subset L_{\emptyset,0}$ which is RE but not recursive. And it contains only (Gödel indexes of) TM machines that always halt.

Of course, other levels of computability can be obtained in the same way.