[Proof] If $x^{2}+x+q_{1}=0$ does not have two distinct real roots, then $\Delta_{1}=1-4 q_{1} \leqslant 0$, i.e., $q_{1} \geqslant \frac{1}{4}$.
In this case, the discriminant of the equation $x^{2}+p x+q_{2}=0$ is
$$\begin{aligned} \Delta_{2} & =p^{2}-4 q_{2}=\left(q_{1}+q_{2}+1\right)^{2}-4 q_{2} \\ & =q_{2}^{2}+2\left(q_{1}+1\right) q_{2}+\left(q_{1}+1\right)^{2}-4 q_{2} \\ & =q_{2}^{2}+2\left(q_{1}-1\right) q_{2}+\left(1+q_{1}\right)^{2} . \end{aligned}$$

The discriminant of the above quadratic trinomial in terms of $q_{2}$ is
$$\begin{aligned} \Delta_{3} & =4\left(q_{1}-1\right)^{2}-4\left(1+q_{1}\right)^{2} \\ & =4\left(-4 q_{1}\right)=-16 q_{1} . \end{aligned}$$

Since $q_{1} \geqslant \frac{1}{4}$, it follows that $\Delta_{3} \leq 0$, i.e., the equation $x^{2}+p x+q_{2}=0$ has two distinct real roots. Therefore, the proposition is proved.