Proof for $\forall x$, $\exists y$ s.t $1/2 \le |x-y|\le 1$ and $ |\cos \pi x - \cos \pi y|\ge 1$ I am trying to understand a certain proof for the indifferentiability of the Weierstrass function, that strongly uses the following lemma:
For all $x \in \mathbb{R}$ there exists $y \in \mathbb{R}$ such that $\frac{1}{2} \le|x-y| \le 1$ and $|\cos \pi x - \cos \pi y|\ge 1$.
As a consequent, it said that for 
$$u_k (x)= a^{-k} \cos\left( \pi b^k x \right)$$ and for $x_0 \in \mathbb {R}$ there exists $x_k$ s.t. 
$$\frac{1}{2} b^{-k} \le |x_k-x_0|\le b^{-k}$$ and $$|u_k (x_k)-u_k(x_0)|\ge a^{-k} \tag{*}$$
I couldn't prove the lemma nor to understand how * follows from it.
Thanks for your help! 
EDIT: I understood how * follows from the lemma, but still unable to prove the lemma.
I tried, using the identity (trig identities don't need proofs :-)) 
$$\cos x - \cos y = -2\sin \frac{1}{2}(x+y)\sin \frac{1}{2}(x-y)$$
to prove that given $x\in[0,\pi]$, there exists $y$ under the above constraints such that $\left|\sin \frac{\pi}{2}(x+y)\right| \left|\sin \frac{\pi}{2}(x-y)\right|\ge\frac{1}{2}$ but couldn't make progress.
 A: By the periodicity of $\cos$, it suffices to look at $x \in [0,2]$.
For $0 \leqslant x \leqslant \frac{1}{3}$, we have $\cos (\pi x) \geqslant \frac{1}{2}$, and therefore
$$\lvert \cos (\pi x) - \cos (\pi(x-1))\rvert = \lvert \cos (\pi x) + \cos (\pi x)\rvert = 2\lvert\cos (\pi x)\rvert \geqslant 1,$$
since $\cos (\alpha+\pi) = -\cos \alpha$. The same computation yields the result for $\frac{2}{3} \leqslant x \leqslant \frac{4}{3}$ and $\frac{5}{3} \leqslant x \leqslant 2$.
For $\frac{1}{3} \leqslant x \leqslant \frac{1}{2}$, we can choose $y = 1$, which yields
$$\lvert \cos (\pi x) - \cos \pi\rvert = \cos x + 1 \geqslant 1,$$
and for $\frac{1}{2} \leqslant x \leqslant \frac{2}{3}$, we can choose $y = 0$. Similarly, we can choose $y = 2$ for $\frac{4}{3}\leqslant x \leqslant \frac{3}{2}$, and $y = 1$ for $\frac{3}{2} \leqslant x \leqslant \frac{5}{3}$.
Not that for $x = k+\frac{1}{2},\, k\in\mathbb{Z}$, there are only two possible choices for $y$, namely $x\pm \frac{1}{2}$, while for $x\in \mathbb{Z}$, all $y\in \left[x-1,x-\frac{1}{2}\right]\cup \left[x+\frac{1}{2},x+1\right]$ work. For $x$ in between, the intervals shrink to single points as $x$ tends to a half-integer.
