I) The proof for $L_1$ is simpler actually; here is an outline:
1) Prove that a linear map $f : E\to E'$ is continuous (even uniformly) iff it is continuous at zero (0); i.e.
$\begin{align} (\exists c\ \epsilon \ \mathbb{R}) \ |f(x)|\leq c|x| \end{align}$
2) Fourier transform is a linear functional defined on $L_1$. So, by (1) you only need to prove it is continuous at 0: We have:
$\begin{align} F(f) = \int_{\mathbb{R}} f(x)e^{-j\omega x}dx \end{align}$ Where F is the Fourier operator defined on L1. $\begin{align} |F(f)| = \left|\int_{\mathbb{R}} f(x)e^{-j\omega x}dx\right| \leq \int_{\mathbb{R}} |f(x)e^{-j\omega x}|dx \leq \int_{\mathbb{R}} |f(x)|dx = \left \| f \right \|_{L^1} < {\infty} \end{align}$ Thus $\begin{align} \left | F(f) \right | \leq 1 \left \| f \right \|_{L^1}, \end{align}$ This completes the proof (set c = 1)
II) Yes, you could use DCT, here's how: Take any sequence $x_n \to 0$ Set $ u_n(x) = f(x)e^{-i\omega (x\pm x_n)} \ \\ \ u(x) = f(x)e^{-i\omega x} $ Clearly, $ u_n \to u $
Now, $\ u\ is\ L^1,\ so\ is\ each\ u_n$
We also have $\left |u_n \right | \leq \left | f \right |\ and\ f\ is\ L^1$ Now use DCT to get: $ \int_{\mathbb{R}}|u_n-u| \to 0\ as \ n \to \infty $