[proofplan] We use the definition of the [quantile function](/page/Quantile%20Function) as the infimum of an upper level set of the distribution function. First we verify that these upper level sets are nonempty and bounded below, using the right-tail and left-tail limits in the definition of a distribution function. Since $p\le q$, every point where $F$ is at least $q$ is also a point where $F$ is at least $p$, so the $q$-upper level set is contained in the $p$-upper level set. Infima reverse inclusions for nonempty bounded-below subsets of $\mathbb R$, giving $F^{-1}(p)\le F^{-1}(q)$. [/proofplan] [step:Define the relevant upper level sets and verify they are nonempty and bounded below] For each $r\in(0,1)$, define the upper level set $A_r\subset\mathbb R$ by \begin{align*} A_r:=\{x\in\mathbb R:F(x)\ge r\}. \end{align*} Since $F$ is a distribution function, $\lim_{x\to\infty}F(x)=1$. Because $r<1$, there exists $x_r\in\mathbb R$ such that $F(x_r)\ge r$, and hence $A_r$ is nonempty. Also $\lim_{x\to -\infty}F(x)=0$. Because $r>0$, there exists $y_r\in\mathbb R$ such that $F(y_r)<r$. Since a distribution function is nondecreasing, every $x\le y_r$ satisfies $F(x)\le F(y_r)<r$, so no such $x$ lies in $A_r$. Thus $y_r$ is a lower bound for $A_r$. Therefore $A_r$ is a nonempty bounded-below subset of $\mathbb R$, so $\inf A_r\in\mathbb R$. By the definition of the quantile function, \begin{align*} F^{-1}(r)=\inf A_r. \end{align*} [/step] [step:Compare the upper level sets and take infima] Since $p\le q$, if $x\in A_q$, then $F(x)\ge q\ge p$, so $x\in A_p$. Therefore \begin{align*} A_q\subseteq A_p. \end{align*} Because $A_q\subseteq A_p$ and both sets are nonempty subsets of $\mathbb R$, every lower bound of $A_p$ is a lower bound of $A_q$. In particular, $\inf A_p\le \inf A_q$. Using the quantile definition from the previous step gives \begin{align*} F^{-1}(p)=\inf A_p\le\inf A_q=F^{-1}(q). \end{align*} This proves the claimed monotonicity. [guided] The order of the argument comes from a simple observation: asking for $F(x)\ge q$ is a stronger requirement than asking for $F(x)\ge p$, because $p\le q$. Let $x\in A_q$. By definition of $A_q$, we have $F(x)\ge q$. Since $q\ge p$, transitivity of the order on $\mathbb R$ gives $F(x)\ge p$, and therefore $x\in A_p$. Hence \begin{align*} A_q\subseteq A_p. \end{align*} Before comparing the infima, we recall from the previous step that $A_p$ and $A_q$ are nonempty bounded-below subsets of $\mathbb R$. This matters because their infima are then [real numbers](/page/Real%20Numbers) and the greatest-lower-bound property applies. Now we translate the set inclusion into an inequality between infima. The larger set is $A_p$, because it is easier for $F(x)$ to exceed the smaller threshold $p$. A larger set can have a smaller infimum. More formally, every lower bound of $A_p$ is automatically a lower bound of its subset $A_q$. Since $\inf A_p$ is the greatest lower bound of $A_p$, it cannot exceed the greatest lower bound of $A_q$. Thus \begin{align*} \inf A_p\le\inf A_q. \end{align*} Finally, the definition of the quantile function identifies these infima with the corresponding quantiles: \begin{align*} F^{-1}(p)=\inf A_p\le\inf A_q=F^{-1}(q). \end{align*} This is exactly the desired inequality. [/guided] [/step]