I just need some hint since I've been stuck in this for several hours..
Let $I_1,I_2,\ldots,I_K$ bne disjoint intervals in $[-1/2,1/2)$,and $f(x)=\sum_{j=1}^K\chi_{I_j}$, where $\chi_I(x)$ is the character function that takes $1$ when $x\in I$ and $0$ if otherwise. Prove that $$\|f(\cdot-y)-f(\cdot)\|_{L^2}\le \sqrt{Ky}$$ Here the inequality is up to constant which is independent of $K,N,y,I_j$. And if $x-y\not \in [-1/2,1/2)$, just translate by integer multiple of $1$ into this region to define $f(x-y)$.
I managed to prove that $\|f(\cdot-y)-f(\cdot)\|_{L^2}\le K\sqrt{y}$(up to constant, the same with all inequality below), this is not hard, since by Parsevel it amounts to dominate $$\sum_{|k|=1}^{\infty}|\hat{f}(k)|^2|e^{2\pi iny}-1|^2\le\sum_{k=1}^\infty\frac{ K^2}{k^2}\sin^2(\pi ny).$$ And discuss the case when $y>1$ and $\frac{1}{k+1}<y\le\frac{1}{k}$.
Also, I could prove that $\|f(\cdot-y)-f(\cdot)\|_{L^2}\le \sqrt{K}$, since $$\sum_{|k|=1}^{\infty}|\hat{f}(k)|^2|e^{2\pi iny}-1|^2\le \sum_{k=1}^\infty(K \sum_{j=1}^K|\hat{\chi}_{I_j}|^2(k))= K \sum_{j=1}^K\|\chi_{I_j}\|_{L^2}^2\le K$$ since $I_j$ are disjoint, and the first inequality is by $|\hat{f}(k)|=|\sum \hat{\chi}_{I_j}(k)|$ and Cauchy-Schwarz.
But I could not prove the required $\sqrt{Ky}$with these techniques. Hope to find some hint, thanks!
Hints (first, read only the first one and try how far you get with that!):
For an arbitrary function $f \in L^1\cap L^2$, you have $$ \Vert f \Vert_2^2 =\int |f|^2\, dx \leq \Vert f \Vert_\infty \cdot \int |f|\, dx =\Vert f\Vert_\infty \Vert f \Vert_1. $$
To obtain an $L^\infty$ bound, use disjointness of the intervals.
For the $L^1$ bound, you can reduce to the case of a single interval using the triangle inequality.