For a set $S\subset \mathbb{N}$, let
$$a(S)=\lim_{n\rightarrow\infty}\frac{\#\{s\in S\>|\>s\le n\}}{n}$$
be the limiting asymptotic density of $S$ in the natural numbers if the limit exists, and $a(S)=0$ if it does not.
Now take $0<\alpha<1$ and let
$$A_\alpha = \{S\subset \mathbb{N}\>|\>a(S)=\alpha\}$$
be the set of all subsets of $\mathbb{N}$ with asymptotic density $\alpha$.
Is $A_\alpha$ countable?
For each $n\in\mathbb N$ choose one element from the set $\{2n-1,2n\}$. Each of the $2^{\aleph_0}$ sets you can get this way has asymptotic density $\frac12$.
P.S. More generally: If $S$ has density $\alpha$ and $T$ has density zero, then the symmetric difference $S\triangle T$ has density $\alpha$. Of course there are continuum many sets of density zero, e.g., the subsets of a given infinite set of density zero. Thus, by fixing $S$ and varying $T$, we get continuum many sets of density $\alpha$.