The Euclid's method to prove that there are infinitely many primes goes as if $p_1,\dots,p_n$ are all the primes, then $p_1\dots p_n+1$ must have a prime divisor which is not among $p_1,\dots,p_n$.
I heard that this very primitive method can yield $\pi(x)\gg \log \log x$ where $\pi(x)=\{p \text{ prime} :p\leq x\}$. But I have no idea how it works.
Here is a variant of this method which yields the bound quickly. The Fermat numbers $F_k=2^{2^k}+1$ are pairwise coprime, since they are odd and satisfy the recursion $F_k-2=F_0\dots F_{k-1}$. So if $P_k$ denotes the least prime divisor of $F_k$, then the primes $P_0,\dots,P_k$ are all distinct. In particular, $\pi(F_k)>k$. From here it follows easily that $\pi(x)\gg\log\log x$.
You can see the same by following Euclid's original method with $p_1\dots p_n-1$ instead of $p_1\dots p_n+1$. Namely, it follows by induction that $p_{k+1}\leq F_k$, whence $\pi(F_k)>k$ again.