\begin{proof} Define the event $E=\{ X\ge\alpha \}=\{ \omega \in\Omega:X(\omega)\ge \alpha \}$. Then $$X\ge \alpha \mathbb{1}_{E}$$where if $\omega \in E$ then $$\begin{cases} X(\omega)\ge \alpha \\ \alpha \mathbb{1}_{E}(\omega)=\alpha \end{cases}\implies X(\omega)\ge \alpha \mathbb{1}_{E}(\omega)$$and if $\omega \not\in E$ then $$\begin{cases} 0\le X(\omega)<\alpha \\ \alpha \mathbb{1}_{E}(\omega)=0 \end{cases}\implies X(\omega)\ge \alpha \mathbb{1}_{E}(\omega)$$hence $$\begin{gather*} \mathbb{E}[X]\ge\alpha \mathbb{E}[\mathbb{1}_{E}]\\ \iff\\ \frac{\mathbb{E}[X]}{\alpha}\ge\mathbb{P}(X\ge \alpha) \end{gather*}$$ \end{proof}