I was wondering if there was another way to derive the inequality shown in the title of this question apart from calculus.
Doing it with calculus is pretty straightforward:
Let $f:(0,1)\to\mathbb R$ be a function defined by $f(p)=p(1-p)$.
Taking the first derivative $f'(p) = 1-2p=0\Rightarrow p=1/2$ is a critical point. Then the maximum is $1/4$ and it is reached when $p=1/2$. Hence $p(1-p)\leq 1/4,$ $\forall p\in(0,1)$.
Another high school solution is given by solving the quadratic equation $4p^2-4p-1=4(p-1)p-1=0$. The zeroes of this polynomial represent the points where the original equation switches from true to false or vice versa, and then simply examining the interval that $(0,1)$ falls into shows that it's true.
We need to show that $4p-4p^2\le 1$ or equivalently that $4p^2-4p+1\ge 0$. But $4p^2-4p+1=(2p-1)^2$.
Substitute $p=\frac{1}{2}+q$. The inequality then becomes
$$\left( \frac{1}{2}+q \right) \left( \frac{1}{2}-q \right)=\frac{1}{4}-q^2 \leq \frac{1}{4}, $$ which is obviously true.
Basic (high school) analytic geometry:
The function $\;y=-x^2+x\;$ is a downwards parabola whose then has at its vertex its maximal point, and the vertex is
$$V=\left(-\frac ba\;,\;\;-\frac{\Delta(=b^2-4ac)}{4a}\right)=\left(-\frac1{-1}\;,\;\;-\frac{1}{-4}\right)=\left(1\,,\,\frac14\right)$$
and thus
$$-x^2+x=x(1-x)\le\frac14\;,\;\;\forall\,x\in \Bbb R$$
