I am looking for the easiest proof that $\binom{n}{k}\leq 2^{n-1}$ (hopefully even with no induction...).
Note: if $k\neq \frac{n}{2}$, then I have a one liner: $$2\binom{n}{k} = \binom{n}{k}+\binom{n}{n-k} \leq (1+1)^n = 2^n.$$
So perhaps it suffices to show that $\binom{2 d}{d} \leq 2^{2d-1}$, but I would prefer a proof with no separation to cases.
$\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n} = 2 \binom{2n-1}{n-1}$
and $\binom{2n-1}{n-1} \le 2^{2n-2}$ by your one liner
$\Rightarrow \binom{2n}{n} \le 2^{2n-1}$