Proving that cosine is uniformly continuous This is what I've already done. Can't think of how to proceed further
$$|\cos(x)-\cos(y)|=\left|-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\leq\left|\frac{x+y}{2}\right||x-y|$$
What should I do next?

The cosine function is Lipschitz-continuous because its derivative is bounded.  That follows from the mean value theorem.  Every Lipschitz-continuous function is uniformly continuous.
postscript prompted by Tom Oldfield's comment: A function $f$ is Lipschitz-continuous iff there is some non-negative number $m$ such that for all $x,y$ in the domain of $f$ we have $|f(x)-f(y)|\le m|x-y|$.  This is stronger than uniform continuity, in that every Lipschitz-continuous function is uniformly continuous, but not every uniformly continuous function is Lipschitz continuous.  An example of a uniformly continuous function that is not Lipschitz-continuous is $x\mapsto\sqrt{1-x^2}$ on the interval $[-1,1]$.  It's really easy to prove that Lipschitz continuity entails uniform continuity.
