variance inequality

Here is a way: Note that $$0 \leq X \leq 1$$ $$\Rightarrow 0\leq X^2 \leq X \leq 1$$ Thus $E[X^2] \leq E[X]$

Now thus $$Var[X] \leq E[X](1-E[X]) $$

But $0 \leq E[X] \leq 1$, the maxima of $f(x)=x(1-x) \quad x\in [0,1]$ is $1/4$ at $x=.5$ (Hint: AMGM).

QED

First, for every real valued random variable $X$ and every real number $x$, $\mathrm{var}(X)\leqslant E[(X-x)^2]$ (and in fact the variance is the minimum over $x$ of these upper bounds). Second, if $X\in[0,1]$ almost surely, then $(X-\frac12)^2\leqslant\frac14$ almost surely.

For $x=\frac12$, these two facts put together yield $\mathrm{var}(X)\leqslant E[(X-\frac12)^2]\leqslant\frac14$.