Upper Bound for $p(1-p) $ where $0 \le p \le 1$

If we do it the AM-GM way instead (valid since both $p, 1-p \ge 0$), we see that

$$\begin{align}\frac{p + (1 - p)}{2} &\ge \sqrt{p(1-p)}\\ \sqrt{p(1-p)} &\le \frac{1}{2}\\ p(1-p) &\le \frac{1}{4}\end{align}$$

with equality at $p = 1-p$ which simplifies into $p = \frac{1}{2}$.

$p(1-p)$ is a parabola with roots at $0$ and $1$. The midpoint of a parabola is where the vertex (maximum if it opens down, minimum if it opens up) occurs, which is at $p=\frac{1}{2}$. In this case, it opens down. You can also just note that if the unconstrained maximum lies within the constraints, then it must be the constrained maximum (that is, argmax(p(1-p)) = 1/2 ignoring the constraints, which lies within $[0,1]$, so it must indeed be the constrained maximum when you constrain to $[0,1]$).

$p(1-p)=p-p^2=-(p^2-p)=-((p-1/2)^2-(1/2)^2)=1/4-(p-1/2)^2 \le 1/4$ (for all values of $p$, but in probability theory only $0 \le p \le 1$ is relevant).