I would like to know if my resolution is right...
I want prove that
$|\varphi(t)| = \vert\mathbb{E}[e^{i t X}]\vert\leq 1$ , $\forall t \in \mathbb{R}$.
$\it{proof:}$
First, note that $|\mathbb{E}[e^{i t X}]| \leq \mathbb{E}[|e^{i t X}|]$.
But, on the other side, we know that $|e^{i t X}| = |\cos(tX)+ i \sin(tX)| = 1$. Then, $\mathbb{E}[|e^{i t X}|] = {\displaystyle\int_{\Omega}1dP} = P(\Omega) = 1$. So,
$|\varphi(t)| = \vert\mathbb{E}[e^{i t X}]\vert\leq \mathbb{E}[|e^{i t X}|] = 1$