Is ${\rm Sym}(X)$ countable if $X$ is countable?
I tried to write all elements in order, like pick the first element as the permutation with the smallest $f(1)$.
If there are several permutations with the same $f(1)$, pick the one with the smallest $f(2)$, or $f(3)$..., then do the same thing to pick the second, the third element and so on.
I wonder if this shows that ${\rm Sym}(X)$ is countable.
No, the symmetric group of an infinite set $X$ is always uncountable, and actually has cardinality $2^{|X|}$.
It is easy to see that it has at least that cardinality, since we can define a surjection $\operatorname{Sym}(X)\to \mathcal{P}(X)$ by sending $f$ to the set of its fixed points. This is surjective since for any subset $Y\subset X$, you can define a symmetry $f$ of $X$ by setting it to the identity on $Y$, and any permutation without fixed points on $X\setminus Y$. (You can try to prove that such a permutation always exist on any set.)