Convergence Issues for Bootstrap Distributions the following is part of a proof from van der Vaarts book on asymptotic statistics:
I want to show that if for a continuous distribution function F
$$P\left(\frac{\hat{\theta}-\theta}{\hat{\sigma}}\leq x\mid P\right) \rightarrow F(x) \mbox{ and } P\left(\frac{\hat{\theta}^\ast-\theta^\ast}{\hat{\sigma}^\ast}\leq x\mid \hat{P}\right)\rightarrow F(x) \text{ (in probability)}$$
$\quad\forall x$
then 
$$\sup_x\| P\left(\frac{\hat{\theta}-\theta}{\hat{\sigma}}\leq x\mid P\right)-P\left(\frac{\hat{\theta}^\ast-\theta^\ast}{\hat{\sigma}^\ast}\leq x\mid \hat{P}\right) \| \rightarrow 0 \text{ (in probability)}$$
So i think it should go like this:
Define $X=\frac{\hat{\theta}_n-\theta}{\hat{\sigma}_n}$ and $\hat{X}=\frac{\hat{\theta}^\ast_n-\hat{\theta}_n}{\hat{\sigma}_n}$.
\begin{align*}
&\sup_x \| P\left(X\leq x\mid P\right)-P\left(\hat{X}\leq x\mid \hat{P}\right) \| \\
\leq&\sup_x\left(\| P\left(X\leq x\mid P\right)-F\|+ \|F-P\left(\hat{X}\leq x\mid \hat{P}\right) \| \right)\\
\leq&\sup_x \| P\left(X\leq x\mid P\right)-F \| +\sup_x \|P\left(\hat{X}\leq x\mid \hat{P}\right)-F\|
\end{align*}
and now i'm a bit lost... i think the claim should follow because the sup is attained because F is continous and monotonous or something like that.
i feel it is close but
 A: Hint: mimic the proof that a tight family of pdf's that converges pointwise to a continuous limit converges uniformly.
(note by OP: german speaking readers may also find a proof of $F_n \rightarrow F$ weakly \Rightarrow \sup \|F_n - F | \rightarrow 0) in Bauer - Mass- und Integrationstheorie)
In my mind, Helly's selection theorem and your problem were related.  
Here is a solution. Let $\epsilon \gt 0$. Since $F$ is continuous with limits at infinity, we can find a partition of $\mathbb{\bar{R}}$ of the form: $a_{0} = -\infty \lt a_{1}\lt\dots \lt a_{k}\lt a_{k+1}=+\infty$ such that
\begin{align*}
& F(a_{i}) \lt F(a_{i-1}) + \epsilon 
\end{align*}
for $i=1,\dots,k+1$.  
We have: 
\begin{align} 
&\sup_x|\hat{F}(x) - \hat{F}^*(x)| = \max_i \sup_{x  \in  [a_{i-1},a_{i})}|\hat{F}(x) - \hat{F}^*(x)| ,
\end{align}
but for $x \in [a_{i-1},a_i)$ we have 
\begin{align}
&-|\hat{F}^*(a_{i-1}) - \hat{F}(a_i)| \le \\  
&\le \hat{F}^*(a_{i-1}) - \hat{F}(a_i) \le\hat{F}^*(x) - \hat{F}(x)\le\hat{F}^*(a_i) - \hat{F}(a_{i-1}) \le \\ 
& \le |\hat{F}^*(a_i) - \hat{F}(a_{i-1})|
\end{align}  
so  
$|\hat{F}^*(x) - \hat{F}(x)|\le max\{|\hat{F}^*(a_i) - \hat{F}(a_{i-1})|,|\hat{F}^*(a_{i-1}) - \hat{F}(a_i)|\}  \\
\stackrel{(a)}{\le} max\{|\hat{F}^*(a_{i-1}) - F(a_{i-1})|, |\hat{F}(a_{i-1} - F(a_{i-1})|,
|\hat{F}^*(a_{i}) - F(a_{i})|, |\hat{F}(a_{i}) - F(a_{i})|,
|F(a_i) - F(a_{i-1})|\} $.
By the selection of $\{a_i\}_i$,  $|F(a_i) - F(a_{i-1})| \le \epsilon $.  
By assumption, for each $i=1,\dots , k$ we can find a value of the of the index that parametrizes the convergence in probability (which is implicit in your statement, and that I will refer to "the sample size") such that beyond that value:
\begin{align*}
& |\hat{F}(a_i) - F(a_i) | \le \epsilon  
\end{align*}
for $i=1, \dots ,k$. Therefore, if the sample size is large enough:
\begin{align} 
&\sup_x|\hat{F}(x) - \hat{F}^*(x)| \stackrel{(b)}{\le} \max_i \{|\hat{F}^*(a_{i}) - F(a_{i})|\} + 2\epsilon .  
\end{align}  
Let $\delta \gt 0$. By the assumption of convergence in probability (as the sample size increases), if the sample size sufficiently large:
\begin{align*}
& P( |\hat{F}^*(a_i) - F(a_i) | \ge \epsilon ) \le \frac{\delta}{k}
\end{align*}
for $i=1, \dots ,k$.  
Finally,
\begin{align*}
& P( \sup_x|\hat{F}(x) - \hat{F}^*(x)| \ge 3\epsilon ) \le \\
& P( \max_i \{|\hat{F}^*(a_{i}) - F(a_{i})|\} + 2\epsilon \ge 3\epsilon ) = \\
& P( \max_i \{|\hat{F}^*(a_{i}) - F(a_{i})|\} \ge \epsilon ) \le \\
& \sum_{i=1}^k P(|\hat{F}^*(a_{i}) - F(a_{i})| \ge \epsilon) \le 
 \sum_{i=1}^k \frac{\delta}{k} = \delta.  
\end{align*}  
Since $\delta \gt 0$ is arbitrary, this shows:
\begin{align*}
& \lim P( \sup_x|\hat{F}(x) - \hat{F}^*(x)| \ge 3\epsilon ) = 0 ,
\end{align*}
and since $\epsilon \gt 0$ is arbitrary, this show:
\begin{align*}
& \sup_x|\hat{F}(x) - \hat{F}^*(x)| \leadsto 0 .
\end{align*}
