First I want to stress that I don't want an answer, perhaps a hint. Let $f(x)$ have period $2\pi$ and let $|f(x) -f(y)| \leq c|x-y|^{\alpha}$, for some constants $c$ and $\alpha$ for all $x$ and $y$. How is that $$|a_n| \leq \frac{c\pi^{\alpha}}{n^{\alpha}}, \qquad |b_n| \leq \frac{c \pi^{\alpha}}{n^{\alpha}}$$
We have $$|a_n| = \frac{1}{\pi} \left |\int_{-\pi}^{\pi}f(x)\cos nx \right| \leq \frac{1}{\pi} \int_{-\pi}^{\pi}|f(x)||\cos nx| dx.$$
I can't apply the Hölder condition to the function $f(x)\cos nx$. Since $f$ has period of $2 \pi$, the upper bound on $|f(x) -f(y)|$ must be $c\pi^{\alpha}$ by selecting the maximum and minimum values of $f$ and noting they must be within $\pi$ of each other. I'm not sure how to link the two.
Summarizing the solution outlined by David C. Ullrich in comments:
Let $f_\delta(t) = f(t-\delta)$, a translation of $f$.
Thanks to the Hölder continuity of $f$, we have $\sup|f-f_\delta|\le C\delta^\alpha$. Hence, for every $n\in\mathbb{Z}$, the Fourier coefficients satisfy $|\widehat{f}(n)-\widehat{f_\delta}(n)|\le C\delta^\alpha$.
On the other hand, $\widehat {f_{\pi/n}}(n) = -\widehat{f}(n)$ because translation by $\pi/n$ amounts to changing the phase of this particular frequency to the opposite one. (In mathematical terms this is the antiperiodicity of cosine and sine.)
Together, the above imply that $|\widehat{f}(n)|\le M(\pi/n)^\alpha$, where $M=C/2$.