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 $c>1$, $$\lim_{n\rightarrow\infty} |S(n,c)|/n = 1,$$