I'm trying to show that pointwise convergence in distribution implies uniform convergence in distribution when the limiting cdf is continuous everywhere using the proof in van der Vaart "Asymptotic Statistics" p. 10. I report here the first part of the proof.
Let $F_n$ and $F$ be the cdf of $X_n$ and $X$. First consider the unidimensional case. Fix $k \in \mathbb{N}$. By continuity of $F$ there exist point $-\infty=x_0<x_1<x_2<...<x_k=+\infty$ with $F(x_i)=\frac{i}{k}$. By monotonicity we have, for $x_{i-1}<x<x_{i+1}$ $$ F_n(x_{i-1})-F(x_{i-1})-\frac{1}{k}=F_n(x_{i-1})-F(x_i) $$ $$ \leq F_{n}(x)-F(x)\leq $$ $$ F_n(x_i)-F(x_{i-1})=F_n(x_i)-F(x_i)+\frac{1}{k} $$ Thus $|F_n(x)-F(x)|$ is bounded above by $sup_{i} |F_n(x_i)-F(x_i)|+\frac{1}{k}$
[...].
Could you help me to clarify the last sentence? I'm confused by the presence of the supremum and the absolute values.
You've shown that for each $x$ there is an $i$ such that $$ F_n(x_{i-1})-F(x_{i-1})-k^{-1}\le F_n(x)-F(x)\le F_n(x_i)-F(x_i)+k^{-1}. $$ Let $M_n$ denote the largest of $|F_n(x_i)-F(x_i)|$ as $i$ ranges from $0$ to $k$; that is, $M_n=\max_{i=0,2,\ldots,k}|F_n(x_i)-F(x_i)|$. (Because we are maximizing over a finite set of numbers, this coincides with $\sup_{i=0,2,\ldots,k}|F_n(x_i)-F(x_i)|$.) In the inequality displayed above, the right side is no more than $M_n+k^{-1}$. Meanwhile, the left side is at least $-M_n-k^{-1}$. Therefore you have shown $$ -M_n-k^{-1}\le F_n(x)-F(x)\le M_n+k^{-1},\qquad\forall x. $$