There are slightly shorter ways for this specific problem (see Ragib's answer and my comment below it), but I will instead push the OP's attempt to completion.
So, we are left with showing two inequalities: $ 2 \sin^2 x - \sin^4 x \geq 0, \tag{1} $ $ 2 \sin^2 x - \sin^4 x \leq 1. \tag{2} $
For (1): An important fact worth remembering is that $0 \leq \sin^2 x \leq 1$ for all real $x$. Multiplying by the nonnegative number $\sin^2 x$, we get $ \sin^4 x \leq \sin^2 x .$
You should be able to prove $(1)$ by plugging in this inequality. Can you complete this part?
For (2): For such inequalities, it is a good idea to collect terms together and see if the resulting expression can be simplified (by grouping terms or factoring). Collecting the terms on the right, the inequality $(2)$ is equivalent to: $ 1 - 2 \sin^2 x + \sin^4 x\geq 0 .\tag{3} $ You should be able to show this by factoring the left hand side. Can you take it from here?