for $T$ an a.s finite stopping time, $B(T-t)-B(T)$ is a brownian motion for $t \in [0,T]$.
I know if $T>0$ is fixed number $B(T-t)-B(T)$ is a brownian motion. To show the claim is true for an a.s stopping time I have done the following.
Let $$T_n = \inf\{k2^{-n} :\: k \in \mathbb{N},\; T \leq k2^{-n}\}$$ the discretization of the stopping time $T$
for $r>0$ let $$B^r(t) := B(r-t)-B(r) \quad \text{ this is a BM}$$ We may write $$B(T_n-t)-B(T_n) = \sum_{k=0}^{\infty} B^{k2^{-n}}(t) \mathbf{1}_{T_n=k2^{-n}}$$
let $s,t \in [0,T]$ $$\mathbb{E}[(B(T_n-t)-B(T_n))B(T_n-s)-B(T_n)]$$ $$ = \sum_{k=0}^{\infty} \mathbb{E}[B^{k2^{-n}}(t)B^{k2^{-n}}(s) \mathbf{1}_{T_n=k2^{-n}}] \stackrel{(1)}{=} s \wedge t \sum_{k=0}^{\infty} \mathbb{P}[T_n=k2^{-n}]=s \wedge t$$ Thus it is a brownian motion (also check $\mathbb{E}[B(T_n-t)-B(T_n)]=0$)
Thus $$s\wedge t = \mathbb{E}[(B(T_n-t)-B(T_n))B(T_n-s)-B(T_n)] \stackrel{(2)}{=} \mathbb{E}[(B(T-t)-B(T))B(T-s)-B(T)]$$
(1) we use $B^{k2^{-n}}(\cdot)$ is BM, is separation of expectation justified?
(2) here I want to use DCT, can we say $B^{T_n}(t):=B(T_n-t)-B(T_n)$, is almost surely bounded on $[0,T]$ by continuity of BM and bound $|B^{T_n}(t)B^{T_n}(s)|\leq C^2$ for some $C>0$?
This is false.
For instance, let $T$ denote the first time $B(t)=0$ after the first time $B(t)=1$. Then e.g. every path of $B(T-t)-B(T)$ hits $1$ before it's negative.