Let $T(n,c)=\{m\in \mathbb{N} \ : \ m<n, \ \omega(m)<(\log\log n)^c\}$. Then we have
$S(n,c)\subset T(n,c)$ and $|S(n,c)|\leq |T(n,c)|$. Also, if
$T'(n,c)=\{ m\in \mathbb{N} \ : \ m<n, \ \omega(m)<(\log\log (\sqrt n))^c\}$, we have
$ |T'(n,c)|+O(\sqrt n) \leq |S(n,c)| \leq |T(n,c)|$.
Thus, we see that if $\lim_{n\rightarrow\infty} |T'(n,c)|/n$ and $\lim_{n\rightarrow \infty} |T(n,c)|/n$ exists and equal,
then the limit $\lim_{n\rightarrow\infty} |S(n,c)|/n$ must exist and equal to
that of $\lim_{n\rightarrow \infty} |T(n,c)|/n=\lim_{n\rightarrow \infty} |T(n,c)|/n$.
As commented above, we use Erdos-Kac Theorem, then we see that $$\lim_{n\rightarrow\infty} |T'(n,c)|/n=\lim_{n\rightarrow \infty} |T(n,c)|/n,$$
and the value of the limit depends on $c$.
For $0<c<1$, the value is $0$,
For $c=1$, the value is $1/2$,
For $1<c$, the value is $1$.
Therefore, we obtain that
If $0<c<1$, $$\lim_{n\rightarrow\infty} |S(n,c)|/n = 0,$$
If $c=1$, $$\lim_{n\rightarrow\infty} |S(n,c)|/n = \frac12,$$
If $c>1$, $$\lim_{n\rightarrow\infty} |S(n,c)|/n = 1,$$