[proofplan]
Let $q=Q(p)$ and trap $Q_n(p)$ inside an arbitrary interval around $q$. The lower bound comes from choosing a continuity point $a<q$ of $F$, which forces $F(a)<p$ and hence eventually $F_n(a)<p$. The upper bound uses continuity of $Q$ at $p$: it gives a level $r>p$ whose quantile is still close to $q$, and then a continuity point $b>Q(r)$ near $q$ with $F(b)\ge r>p$. Applying the [inverse inequalities for generalized quantiles](/theorems/8346) to $F_n$ at the two continuity points $a$ and $b$ gives $a<Q_n(p)\le b$ for all large $n$, and the arbitrariness of the interval yields convergence.
[/proofplan]
[step:Record density of continuity points for the limiting distribution function]
We first note that the continuity points of $F$ are dense in $\mathbb R$.
[claim:Continuity points of a monotone real function are dense]
Let $G:\mathbb R\to\mathbb R$ be nondecreasing. Then every nonempty open interval in $\mathbb R$ contains a continuity point of $G$.
[/claim]
[proof]
Let $I\subset\mathbb R$ be a nonempty open interval. For $m\in\mathbb N$, define
\begin{align*}
D_m:=\{x\in I:G(x+)-G(x-)>\frac{1}{m}\},
\end{align*}
where
\begin{align*}
G(x-):=\lim_{y\uparrow x}G(y),\qquad G(x+):=\lim_{y\downarrow x}G(y).
\end{align*}
These one-sided limits exist in $\mathbb R\cup\{-\infty,\infty\}$ by monotonicity. Fix a compact interval $J=[\alpha,\beta]\subset I$. If $x_1<\cdots<x_k$ are points of $D_m\cap J$, choose numbers
\begin{align*}\alpha< u_1<x_1<v_1<u_2<x_2<v_2<\cdots<u_k<x_k<v_k<\beta.\end{align*}
By monotonicity,
\begin{align*}\sum_{i=1}^{k}\bigl(G(v_i)-G(u_i)\bigr)\le G(\beta)-G(\alpha).\end{align*}
Letting $u_i\uparrow x_i$ and $v_i\downarrow x_i$ for each $i$ gives
\begin{align*}\frac{k}{m}<\sum_{i=1}^{k}\bigl(G(x_i+)-G(x_i-)\bigr)\le G(\beta)-G(\alpha).\end{align*}
Hence $D_m\cap J$ is finite. Since $I$ is the union of countably many compact intervals contained in $I$, each $D_m$ is countable. The set of discontinuities of $G$ in $I$ is contained in
\begin{align*}
\bigcup_{m=1}^{\infty}D_m,
\end{align*}
because a discontinuity of a monotone function has a positive jump. Thus the discontinuities in $I$ are countable. Since $I$ is uncountable, $I$ contains at least one continuity point of $G$.
[/proof]
Applying the claim to the nondecreasing distribution function $F$ shows that every nonempty open interval contains a continuity point of $F$.
[/step]
[step:Choose a continuity point to the left of the target quantile]
Let
\begin{align*}
q:=Q(p).
\end{align*}
Fix $\varepsilon>0$. Since $p\in(0,1)$ and $F$ is a distribution function, the limits $F(x)\to0$ as $x\to-\infty$ and $F(x)\to1$ as $x\to\infty$ imply that the set $\{x\in\mathbb R:F(x)\ge p\}$ is nonempty and bounded below; hence $q\in\mathbb R$. By the density established above, choose a continuity point $a\in(q-\varepsilon,q)$ of $F$. Since $F$ is a distribution function, $p\in(0,1)$, and $a<q=Q(p)$, the inverse inequality for quantiles [citetheorem:8346] applied to $F$ gives
\begin{align*}
F(a)<p.
\end{align*}
Because $a$ is a continuity point of $F$ and $F_n(a)\to F(a)$, there exists $N_1\in\mathbb N$ such that for every $n\ge N_1$,
\begin{align*}
F_n(a)<p.
\end{align*}
Since $F_n$ is a distribution function and $p\in(0,1)$, applying [citetheorem:8346] to $F_n$ gives
\begin{align*}
a<Q_n(p)
\end{align*}
for every $n\ge N_1$.
[/step]
[step:Use continuity of the limiting quantile to find a strict upper comparison point]
Since $Q$ is continuous at $p$, there exists $\delta>0$ such that
\begin{align*}
0<\delta<1-p
\end{align*}
and such that for every $r\in(0,1)$ with $|r-p|<\delta$,
\begin{align*}
|Q(r)-q|<\frac{\varepsilon}{2}.
\end{align*}
Define
\begin{align*}
r:=p+\frac{\delta}{2}.
\end{align*}
Then $r\in(p,1)$ and
\begin{align*}
Q(r)<q+\frac{\varepsilon}{2}.
\end{align*}
By density of the continuity points of $F$, choose a continuity point $b$ of $F$ satisfying
\begin{align*}
Q(r)<b<q+\varepsilon.
\end{align*}
Since $F$ is a distribution function, $r\in(0,1)$, and $Q(r)<b$, the inverse inequality [citetheorem:8346] applied to $F$ at level $r$ gives
\begin{align*}
r\le F(b).
\end{align*}
Since $r>p$, this implies
\begin{align*}
p<F(b).
\end{align*}
[guided]
The right-hand comparison is the only place where continuity of $Q$ at $p$ is used. We need a point $b$ just to the right of $q$ with the strict inequality $F(b)>p$. Such a point need not exist merely from the definition of $q=Q(p)$, because $F$ could be flat at height $p$ immediately to the right of $q$. Continuity of $Q$ at $p$ rules out that obstruction.
Since $Q$ is continuous at $p$, we may choose $\delta>0$ with $0<\delta<1-p$ such that
\begin{align*}|Q(r_0)-q|<\frac{\varepsilon}{2}\end{align*}
whenever $r_0\in(0,1)$ and $|r_0-p|<\delta$. Define
\begin{align*}
r:=p+\frac{\delta}{2}.
\end{align*}
Then $r\in(p,1)$, and continuity gives
\begin{align*}
Q(r)<q+\frac{\varepsilon}{2}.
\end{align*}
We now choose $b$ not merely with $b>q$, but with $b>Q(r)$. The earlier density claim applies because $F$ is nondecreasing as a distribution function, so every nonempty open interval contains a continuity point of $F$. Since $Q(r)<q+\varepsilon/2<q+\varepsilon$, the interval $(Q(r),q+\varepsilon)$ is nonempty, and therefore there is a continuity point $b$ of $F$ such that
\begin{align*}
Q(r)<b<q+\varepsilon.
\end{align*}
Because $F$ is a distribution function, $r\in(0,1)$, and $Q(r)<b$, the generalized inverse characterization [citetheorem:8346], applied to $F$ at level $r$, yields
\begin{align*}
r\le F(b).
\end{align*}
Finally $r=p+\delta/2>p$, so
\begin{align*}
p<F(b).
\end{align*}
Thus $b$ is a continuity point of $F$ lying within $\varepsilon$ of $q$ on the right and satisfying the strict upper comparison inequality needed for the approximating distribution functions.
[/guided]
[/step]
[step:Transfer the two strict inequalities to the approximating distribution functions]
Since $b$ is a continuity point of $F$ and $F_n(b)\to F(b)$, while $p<F(b)$, there exists $N_2\in\mathbb N$ such that for every $n\ge N_2$,
\begin{align*}
p<F_n(b).
\end{align*}
In particular,
\begin{align*}
p\le F_n(b).
\end{align*}
Since $F_n$ is a distribution function and $p\in(0,1)$, applying the inverse inequality [citetheorem:8346] to $F_n$ gives
\begin{align*}
Q_n(p)\le b
\end{align*}
for every $n\ge N_2$.
[/step]
[step:Trap the approximating quantiles and let the interval shrink]
Let
\begin{align*}
N:=\max\{N_1,N_2\}.
\end{align*}
For every $n\ge N$, the preceding two bounds give
\begin{align*}
a<Q_n(p)\le b.
\end{align*}
Using the choices $a\in(q-\varepsilon,q)$ and $b\in(Q(r),q+\varepsilon)$, we obtain
\begin{align*}
q-\varepsilon<Q_n(p)<q+\varepsilon
\end{align*}
for every $n\ge N$. Therefore
\begin{align*}
|Q_n(p)-q|<\varepsilon
\end{align*}
for every $n\ge N$. Since $\varepsilon>0$ was arbitrary and $q=Q(p)$, this proves
\begin{align*}
\lim_{n\to\infty}Q_n(p)=Q(p).
\end{align*}
[/step]
