The proof that I have in mind is as follows - 


$\text{Gal }(\overline{\mathbb Q}/\mathbb Q)$ is a proper uncountable subgroup of the group of permutations on countably many symbols, hence the latter is uncountable. .


But it needs a lot of jargon from topology and algebra. Is there a neat proof like Cantor's diagonal argument?



An easy proof of the uncountability of bijections on natural numbers?
Indeed, Cantor's argument is readily adapted. Let $\pi\_1,\pi\_2,\pi\_3, \ldots $ be any countable sequence of permutations of $\mathbb N$ ; let us show that this sequence
does not exhaust all permutations, by constructing a permutation $\pi$ different from all 
the $\pi\_i$. We first define $\pi$ on the even integers inductively, then define
$\pi$ on the odd integers.


Let $X$ be any subset of $\mathbb N$ such that both $X$ and ${\mathbb N} \setminus X$
are infinite (e.g. the even integers, the prime numbers ...).  

 Set $\pi(0)$ to be an integer in $X$ different from $\pi\_1(0)$. Set $\pi(2)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi\_2(2)\rbrace$. Set $\pi(4)$ to be an integer in $X$ not in $\lbrace \pi(0),\pi(2)\pi\_3(4)\rbrace$. Continuing like this inductively, we define an injection from the even integers to $X$,
such that $\pi(2k-2) \neq \pi\_k(2k-2)$ for any $k$ (and hence $\pi \neq \pi\_k$).


Finally, the set A=${\mathbb N} \setminus \pi(2\mathbb N)$ is countably infinite ; setting
$\pi(2k-1)=$ the $k$-th element of $A$ finishes the proof. 


