Let $f$ be of moderate decrease, i.e., there exists $A>0$ such that $f(x)\le \frac{A}{1+x^2}$ for all $x$. Then we can define the Fourier Transform by $$\hat{f}(\xi):= \int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}dx.$$ Of course, $|e^{-2\pi ix\xi}|=1$ so the integrand is of moderate decrease and the integral makes sense. In fact, this last observation implies that $\hat{f}$ is bounded.
I'm trying to show this part. I know that we have
$$\left|\int_{-N}^Nf(x)e^{-2\pi ix\xi} dx\right|\le 2N\frac{A}{1+x^2}$$
because the integrand is of moderate decrease, but if we make $N\to\infty$ to get $\hat f$ on the LHS, the RHS also goes to infinity, so I don't know how to use the fact of the integrand being of moderate decrease to show the boundedness of $\hat f$.
$$|\hat{f}(\xi)| \le \int_{-\infty}^\infty |f(x)| dx \le A \int_{-\infty}^\infty \frac{1}{1+x^2} dx \:.$$ From this you get the result because the integral on the right hand side is finite.