(Proof verification): $f(x) = $ sin$^2x$ is uniformly continuous of $\mathbb{R}$ My proof: The theorem I am trying to apply is, "continous function on  a compact set is uniformly  continous". Since $\mathbb{R}$ is not compact, we cannot directly apply the above theorem. But we notice that since sin$^2x$ is periodic all I need is the compact domain $[n\pi,(n+2)\pi]$ because by varying $n$, I can cover $\mathbb{R}$. So I do the following, if given $\mid x - y \mid < \delta$, then I fit in $x,y \in [n\pi, n+2)\pi]$ for some $n$. Since $f$ is uniformly continuos in this interval, I will have my $\epsilon$. 
Example of my explanation: Let $x = 1.81\pi$ and $= 1.9\pi$, then $\mid x - y\mid < 0.1\pi$ and $x,y \in [0, 2\pi]$, so we can find an $\epsilon$ because $f$ is uniformly continuos in $[0, 2\pi]$. The only way this process could break is when $x = 1.95\pi$ and $y = 2.04\pi$, then $\mid x - y \mid < 0.1\pi$ but $x, y \not\in [0, 2\pi]$ so we cannot find the appropriate $\epsilon$. But here I make a clever change of domain to $[\pi, 3\pi]$ from $[0,2\pi]$. In $[\pi, 3\pi]$ I can use the previous reasoning to find the $\epsilon$.
Is my proof correct?
 A: The independence of $\delta$ on location ($n$ in your paradigm) is a consequence of periodicity, but requires some additional argument.  You can fix your proof as follows.  
As you observed,  we have uniform continuity on the compact interval $[0,4\pi].$ Thus, for any $\epsilon > 0$ there exists $\delta >0$ such that $|\sin^2 x - \sin^2 y| < \epsilon$ for all $x,y \in [0,4\pi]$ when $|x-y| < \delta$.
For all $x,y \in \mathbb{R}$, if $|x-y| < \delta' = \min(\delta,2\pi)$ then $x,y \in [z, z+2 \pi]$ for some $z \in \mathbb{R}$.  But $z  = 2m\pi + \alpha$ for some integer $m$ and $0 \leqslant \alpha < 2 \pi$. Hence, 
$$|\sin^2 x - \sin^2 y| = |\sin^2(2m\pi+ \alpha + x' )- \sin^2(2m\pi + \alpha + y')| \\= |\sin^2( \alpha +x' )- \sin^2( \alpha +y')| $$
where $x' = x-z$, $y' = y-z,$ and $x',y' \in [0,2 \pi]$
Since $|\alpha + x' - (\alpha+y')| = |x-y| < \delta$ and $\alpha +x', \alpha + y' \in [0,4\pi]$, we have
$$|\sin^2 x - \sin^2 y| = |\sin^2( \alpha +x' )- \sin^2( \alpha +y')| < \epsilon$$
for all $x,y$ such that $|x- y| < \delta' =\min(\delta,2\pi)$.
Since $\delta'$ does not depend upon $x$ and $y$, we have uniform continuity.
