I want to prove the following:
Let $f \in C^k(\mathbb{R})$ and $\mathcal{F}$ denote the Fourier transformation. If $f$ has compact support, then there exists some $C \in \mathbb{R}$ such that $\left|(\mathcal{F}f)(x)\right| \leq \frac{C}{1+|x|^k}$
Here is my attempt: I know that one property of the Fourier transformation is the following: $(\mathcal{F}f')(\xi)=2 \pi i\, \xi \,(\mathcal{F}f)(\xi)$.
Using the mentioned property I get $$(1+|x|^k)\,\left|(\mathcal{F}f)(x)\right| = \left|(\mathcal{F}f)(x)\right|+ \left|x^k\, (\mathcal{F}f)(x)\right| \\ = \left|(\mathcal{F}f)(x)\right|+ \tfrac{1}{(2 \pi)^{k}} \left|(\mathcal{F}f^{(k)})(x)\right| \\ = \left|\frac{1}{\sqrt{2 \pi}}\int_{\mathbb{R}} f(\xi)\,e^{-ix \xi} \,\mathrm d \xi\right| + \tfrac{1}{(2\pi)^{k}}\left|\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} f^{(k)}(\xi)\,e^{-ix \xi}\,\mathrm d \xi\right| \\ \leq \int_{\mathbb{R}} |f(\xi)| \,\mathrm d\xi +\int_{\mathbb{R}} |f^{(k)}(\xi)| \,\mathrm d\xi $$ Since $f$ has compact support, the above is bounded by $C := |\mathrm{supp}(f)| \,(\mathrm{sup}(f) + \mathrm{sup}(f^{(k)}))$, and so $$ |\mathcal{F}f(x)| \leq \frac{C}{1+|x|^k} $$
Is my calculation/argumentation correct? Is there a similar inequality for non-differentiable functions?
Your computation is correct. More generally, as you computation shows, you do not need compact support of $k$ bounded derivatives, you just need $f\in L^1$ (i.e. integrable) and $f^{(k)}\in L^1$, that is $f\in W^{k,1}$, the Sobolev space of functions verifying exactly these two conditions. The fact that this is true for such functions even if $f^{(k)}$ is not continuous and even if $f$ is not compactly supported follows by density of smooth and compactly supported functions in $W^{k,1}$.
Even more generally, the result also holds if $f^{(k)}$ is a bounded Radon measure. If $k=1$, that would mean that it is sufficient to have $f\in L^1\cap BV$, where $BV$ is the space of functions with bounded variations. As an example, the Fourier transform of $f(x) = \mathbf 1_{[-1,1]}$ is the sine cardinal function $\mathrm{sinc}$ which indeed verifies $\mathrm{sinc}(x) \leq \frac{C}{1+|x|}$.