Riemann-Lebesgue lemma:
Let ${\displaystyle f\in L^{1}(\mathbb {R} ^{n})}$ be an integrable function, i.e. ${\displaystyle f\colon \mathbb {R} ^{n}\rightarrow \mathbb {C} }$ and let ${\hat {f}}$ be the Fourier transform of $f$ i.e. $${\displaystyle {\hat {f}}\colon \mathbb {R} ^{n}\rightarrow \mathbb {C} ,\ \omega \mapsto \int _{\mathbb {R} ^{n}}f(x)\mathrm {e} ^{-\mathrm {i} x\cdot \omega }\mathrm {d} x.}$$
Then ${\hat {f}}$ vanishes at infinity: ${\displaystyle |{\hat {f}}(\xi )|\to 0}$ as ${\displaystyle |\xi |\to \infty }$.
Proof
$${\displaystyle {\hat {f}}(\omega )=\int _{\mathbb {R} }f(x)\mathrm {e} ^{-\mathrm {i} x\omega }\mathrm {d}x}$$
Supstituting ${\displaystyle \textstyle x\to x+{\frac {\pi }{\omega }}}$
$${\displaystyle {\hat {f}}(\omega )} = {\int _{\mathbb {R} }f\left(x+{\frac {\pi }{\omega }}\right)\mathrm {e} ^{-\mathrm {i} x\omega }\mathrm {e} ^{-\mathrm {i} \pi }\mathrm {d} x=-\int _{\mathbb {R} }f\left(x+{\frac {\pi }{\omega }}\right)\mathrm {e} ^{-\mathrm {i} x\omega }\mathrm {d} x}$$
Than I use the triangle inequality for integrals:
$$|{\displaystyle {\hat {f}}(\omega )}| \leq {-\int _{\mathbb {R} }|f\left(x+{\frac {\pi }{\omega }}\right)\mathrm {e} ^{-\mathrm {i} x\omega }\mathrm| {d} x}$$
$$|{\displaystyle {\hat {f}}(\omega )}| \leq {-\int _{\mathbb {R} }|f\left(x+{\frac {\pi }{\omega }}\right)\mathrm | {d} x}$$
Also:
$$|{\displaystyle {\hat {f}}(\omega )|\leq\int _{\mathbb {R} }|f(x)\mathrm {e} ^{-\mathrm {i} x\omega }\mathrm| {d}x}$$
$$|{\displaystyle {\hat {f}}(\omega )|\leq\int _{\mathbb {R} }|f(x)\mathrm | {d}x}$$
Taking the mean of the two expressions we get:
$$|{\displaystyle {\hat {f}}(\omega )|\leq \frac{1}{2}\int _{\mathbb {R} }|f(x)\mathrm {-f\left(x+{\frac {\pi }{\omega }}\right)\mathrm|{d}x}}$$
Now, becasue $f$ is continous ${\displaystyle \left|f(x)-f\left(x+{\tfrac {\pi }{\omega }}\right)\right|}$ converges to $0$ as ${\displaystyle |\omega |\to \infty }$ for all $x\in \mathbb{R} $.
Thus, ${\displaystyle |{\hat {f}}(\omega )|}$ converges to $0$ as ${\displaystyle |\omega|\to \infty }$ due to the dominated convergence theorem.
Is this proof ok? How can $|{\displaystyle {\hat {f}}(\omega )}|$ be less than ${-\int _{\mathbb {R} }|f\left(x+{\frac {\pi }{\omega }}\right)\mathrm | {d} x}$ if $|{\displaystyle {\hat {f}}(\omega )}|$ is in absolute value?
If $\hat{f}(\omega) = - \int_\mathbb{R}f\left(x+\frac{\pi}{\omega}\right)e^{-ix\omega}\,\text{d}x$, taking the absolute value and using the triangle inequality for integrals yields $\vert\hat{f}(\omega) \vert\leq \int_\mathbb{R}\left\vert f\left(x+\frac{\pi}{\omega}\right)\right\vert\,\text{d}x$. So you don't have the minus sign there anymore due to the absolute value. Additionally you forgot the absolute value in your last and second to last displayed equations on the left-hand side. Otherwise your proof looks fine to me.
Edit: While the minus there is wrong due to the absolute value, I do understand the problem. But there is an easy way to fix it, we can just take the absolute value a step later: \begin{align*} \hat{f}(\omega) &= \frac{1}{2}\hat{f}(\omega) + \frac{1}{2}\hat{f}(\omega) = \frac{1}{2} \int_\mathbb{R} f(x)e^{-ix\omega} \,\text{d}x - \frac{1}{2}\int_\mathbb{R} f\left(x+\frac{\pi}{\omega}\right)e^{-ix\omega} \,\text{d}x \\ &= \frac{1}{2} \int_\mathbb{R} \left[f(x) - f\left(x+\frac{\pi}{\omega}\right)\right]e^{-ix\omega}\,\text{d}x. \end{align*} Now you can take absolute values and conclude as above.