Please help me with this (source and context follows after the question). Thank you!
Question: Let $F_1,\ldots,F_n,\ldots$ and $F$ be distribution functions with corresponding quantiles $F_1^*,\ldots,F_n^*,\ldots$ and $F^*$. Suppose that $$ \lim_{n\to\infty}F_n(x)=F(x)\quad\text{for all continuity points }x\text{ of }F. $$
Suppose that $u\in(0,1)$ and that $y$ is a continuity point of $F$ with $F^*(u+)<y$. Show that $F^*_n(u)\leq y$ for all large $n$.
Source and context: I'm doing a review on probability theory using https://galton.uchicago.edu/~wichura/stat304.html. The above question appears as part (b) of Exercise 10 on the notes on distribution functions. Part (a) is as follows:
(a) Suppose that $u\in(0,1)$ and that $w$ is a continuity point of $F$ with $F^*(u)>w$. Use the switching formula to show that $F_n^*(u)>w$ for all large n.
Solution: the switching formula referred above is: $$ u\leq F(x)\iff F^*(u)\leq x\quad(\text{alternative form: } u>F(x)\iff F^*(u)>x). $$ Using this formula in the 1st and 3rd implications below for large $n$ gives part (a): $$ F^*(u)>w\implies u>F(w)\implies u>F_n(w)\implies F_n^*(u)>w. $$
I think I have got it!
Let $u_m=u+\frac{1}{m}$ where $m$ is an integer large enough so that $u+\frac{1}{m}<1$. Because $F^*(u+)<y$, we can make $m$ large enough so that $F^*(u_m)<y$. The latter, thanks to the Switching Formula, means $F(y)\geq u_m>u$. The last strict inequality is important because it allows us to infer that for all $n$ sufficiently large, $u\leq F_n(y)$. Using the Switching Formula one last time, we obtain that $F_n^*(u)\leq y$ for all large $n$.
Thanks for your time!