Is $\sin^2x$ uniformly continuous on$x\in [0,\infty]$

I'm not sure if you are "allowed" to use this method, but for a place to start:

Note that the derivative $2\sin x\cos x$ is bounded between $[-2,2]$

Now we want to show that for any $\epsilon$, we can pick $\delta$ such that $|x-y|<\delta$ implies $|\sin^2x-\sin^2y|<\epsilon$.

And since the derivative is bounded, we have that for all $x,y$ $$\left|\frac{\sin^2x-\sin^2y}{x-y}\right|\le 2$$

And you can probably take it from here.

Carrying your idea further, we have for $|x-y| < \delta(\epsilon) = \epsilon/2$

$$|\sin^2 x - \sin^2 y|= |\sin x + \sin y||\sin x - \sin y| \\ \leqslant 2 |\sin x - \sin y|\\ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right| \\ \leqslant 4\left|\sin\left(\frac{x-y}{2}\right)\right|\\ \leqslant 2|x-y| \\< \epsilon.$$