= Solution
The <Prime number theorem> says $\pi(x)\sim x/\log x$ as $x\to\infty$. <Bertrand's postulate> says that for every integer $n\ge2$ there is a <prime number> strictly between $n$ and $2n$.
To count <S-smooth numbers>, split each prime exponent into its parity and an even part. This gives the unique expression $n=ab^2$, where $a$ is a <squarefree> product of a subset of $S$. There are $2^{|S|}$ possible $a$, while $b\le\sqrt x$ whenever $n\le x$. The <squarefree-part bound for smooth numbers> therefore gives $f_S(x)\le2^{|S|}\lfloor\sqrt x\rfloor\le2^{|S|}\sqrt x$.
For an integer $x\ge1$, take $S$ to be all <prime numbers> at most $x$. Every positive integer at most $x$ is then <S-smooth>, so $x\le2^{\pi(x)}\sqrt x$. Taking <logarithms> yields
$$
\boxed{\pi(x)\ge\frac{\log x}{2\log2}.}
$$
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