Let $\sum =\left \{ 0,1 \right \}$ be alphabet.
$\sum^{*}$= Set of all strings from alphabet
L=set of all languages , It is actually power set of $\sum^{*}$ .
now $\sum^{*}$ is countable set as there can be one-to-one corresponding between $\sum^{*}$ and natural number.
Now L is an uncountable set as from cantor theorem, if S is Countably infinite set then P(S) is uncountable. Here $\sum^{*}$ is countable set so it’s power set is L so ,L is uncountable.
So set of all language is uncountable.
now , we know diagonalization language is not Recursively enumerable language(not RE) but still it is a language which is subset of $\sum^{*}$ .As $\sum^{*}$ is countable and subset of every countable set is also countable so Diagonalization language is also countable .
why set of all non RE language is uncountable(not related to question) ?:-
Now,
set of all language ={set of all RE language} $\cup$ {set of all Non RE language}
Now , We know that set of all RE language is countably infinite as there exist a TM for an RE language and set of turing machine is countable.
Now as the Set of all RE language is countable , now to make the set of all language uncountable , the set of Non RE language must be uncountable.
So set of all non recursive enumerable language is uncountable .
https://math.stackexchange.com/questions/551630/subset-of-a-countable-set-is-itself-countable
https://en.wikipedia.org/wiki/Cantor%27s_theorem
https://gateoverflow.in/378523/self-doubt?state=edit-378531