Define $$ Y = \begin{cases} d & X \ge d \\ 0 & \text{otherwise} \\ \end{cases}. $$ Then $0 \le Y \le X$ Additionally, it follows immediately from the definition that $Y$ is a random variable (i.e., that it is measurable). Computing the expected value of $Y$ we have that $$ \mathbb{E}[X] \ge \mathbb{E}[Y] = d\cdot\mathbb{P}_{}\left\{X \ge d\right\}, $$ and the inequality follows.