What is the simplest to prove, non-trivial upper bound of prime counting function? The thread here lists some pretty unique, simple, and non-trivial proofs on lower bounds for the prime counting function. We can also ask the analogous question: 

What is the simplest to prove, non-trivial upper bound for the prime
  counting function ?

I consider trivial to be something along the lines of $\pi(n) \le n/2$  or worse bounds. 

Chebyshev's theorem is a weak version of the PNT, and the upper bound
$$ \forall x\geq x_0,\qquad \pi(x)\leq (2\log 2)\cdot\frac{x}{\log x} \tag{1}$$
is not so difficult to prove with a subtle argument. From Stirling's approximation/Wallis product we have that the central binomial coefficient behaves in the following way:
$$ C_n=\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{2} $$
but every prime in the range $(n,2n)$ is a divisor of $C_n$, hence
$$ \sum_{n<p<2n}\log p \leq \log C_n \leq n\log 4\tag{3} $$
and by considering the intervals $(n,2n)$, $\left(\frac{n}{2},n\right)$,$\left(\frac{n}{4},\frac{n}{2}\right)$ and so on we get:
$$ \vartheta(x) = \sum_{p<x}\log p \leq x\log 4.\tag{4} $$
Now it is enough to apply summation by parts to get $(1)$.
