Prove that $f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, -1) < B_{\tau_{n-1 }} < f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, 1)$ 
I would like to prove that (almost surely) 
$$f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, -1) < B_{\tau_{n-1 }} < f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, 1)$$

Where the context is as follows: we have a martingale $(X_n )$ - with expectation equal to zero - such that for each $n \ge 1$ there exists a Borel measurable function $f_n: \ \mathbb {R } ^{-1 } \times \{-1, 1 \}  \to \mathbb R$, and a $\{-1 , 1 \} $- valued random variable $D_n $ such that 
$$X_n = f_n(X_1, \dots , X_{n-1 } , D_n )$$
Further we assume that for any $x_1, \dots, x_{n-1 } $
$$f_n(x_1, \dots, x_{n-1 } , -1 ) < f_n(x_1, \dots, x_{n-1 } , 1 )$$

Given a Brownian motion $(B_t)$ we define the stopping times $\tau_0 = 0$ and for $n \ge 1$
$$\tau_n = \inf \{t > \tau_{n-1 }: \ B_t \in \{f_n(B_{\tau_1 } ,\dots, B_{\tau_{n-1 } }, -1 ), f_n(B_{\tau_1 } ,\dots, B_{\tau_{n-1 } }, 1 ) \} \}   $$


This is what I manage to do:
For $n=1 $ we have that $f_1 : \{-1, 1 \} \to \mathbb{R}$, and since $f_1(D_1)=X_1 $ and $E[X_1]=E[X_0]=0$, we get
$$0 = E[f_1(D_1)]=f_1(-1)P[D_1=-1] + f_1(d)P[D_1=1]$$
By the assumption that $f_1(-1) < f_1(1)$ this means that $f(-1)<0<f_1(1)$. And since $\tau_0 = 0$ and $B_0 = 0$ the claim holds for $n=1$.
For general $n $ we have again that
$$E[f_n(X_1, \dots, X_{n-1 } , D_n )] = 0$$
And hence
\begin{multline*}
    E[f_n(X_1, \dots, X_{n-1 } , D_n )] = \\
    = E[f_n(X_1, \dots, X_{n-1 } , -1 )1_{\{D_n = -1\}}] + E[f_n(X_1, \dots, X_{n-1 } , 1 )1_{\{D_n = 1\}}] = 0
\end{multline*}
This means that one of the integrals must be negative and one positive [or both equal to zero]. But here I got stuck!

How is it possible to relate the value of $f_n(B_{\tau_1 } , \dots B_{\tau_{n-1 } } , \pm 1 ) $ to $B_{\tau_{n-1 } } $?

Much grateful for any help provided!
 A: Combining the martingale property with the inequality $f_n(x_1, \dots, x_{n-1 } , -1 ) < f_n(x_1, \dots, x_{n-1 } , 1 )$ and assuming that $\mathrm P(D_n = 1\mid \mathcal F_{n-1})\notin \{0,1\}$ a.s., we have
$$
X_{n-1} = \mathrm E[X_n \mid \mathcal F_{n-1}] = \mathrm E[f_n(X_1,\dots,X_{n-1},D_n) \mid \mathcal F_{n-1}] \\
= f_n(X_1,\dots,X_{n-1},1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1}) \\+ f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1})\\
< f_n(X_1,\dots,X_{n-1},1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1})\\ + f_n(X_1,\dots,X_{n-1},{\color{red}1})\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1})\\
= f_n(X_1,\dots,X_{n-1},1).
$$
Similarly, 
$$
X_{n-1} > f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1})\\ + f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1})
= f_n(X_1,\dots,X_{n-1},-1).
$$
A: Following zhoraster's answer I felt that the last step was missing. At least it wansn't obvious to me. So below is what I would try to do. I'm sorry for the argument being extremely long winded. Hopefully there is a better way than this!
That is I would like to show that 
\begin{equation}
f(X_1, \dots, X_{n-1 },-1 ) < X_{n-1 } < f(X_1, \dots, X_{n-1 }, 1)
\end{equation}
implies
$$f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, -1) < B_{\tau_{n-1 }} < f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, 1)$$
For $n=2 $  we would have $ \tau_{1 } = \inf \{t>0: B_t \in \{f_1(-1), f_1(1) \} \}$. Thus $B_{\tau_1 } $ equals either $f_1(-1) $ or $f_1(1)$ that is $\tau_1 $ may only take two different values.
For 
$$f_2(B_{\tau_1 }, -1) < B_{\tau_1}  < f_n(B_{\tau_1 }, 1)$$ 
to hold it should thus be the case that 
$$f_2(f_1(-1), -1 ) < f_1(-1) < f_2(f_1(-1) , 1)$$
and 
$$f_2(f_1(1), -1 ) < f_1(1) < f_2(f_1(1) , 1)$$
If we take $\omega$ such that $X_1( \omega) = f(-1)$, then the first inequality above follows from (1), and similarly do the second above for an $\omega $ such that $X_1(\omega) = f_1(1)$.
For $n=3$ we would have $\tau_2 = \inf \{t > \tau_1 : B_t \in \{f_2(B_{\tau_1 } ,-1), f_2(B_{\tau_1 } ,1 ) \} \} $
and we should show that 
$$f_3(B_{\tau_1 }, B_{\tau_2 } , -1 ) < B_{\tau_2 } < f_2(B_{\tau_1 }, B_{\tau_2} ,1)$$
Thus given $\omega $ we must consider first what values $B_{\tau_1 } $ and $B_{\tau_2 }$  may take and then consider if the two inequalities hold for those values. As shown in the previous step $\tau_1 $ may only take the values $f_1(-1) $ or $f_1(1) $.  Thus $\tau_2 $ may be written
$$ 
\tau_2 =
\begin{cases}
\inf \{t > f_1(-1): B_t \in \{f_2(f_1(-1), -1), \ f_2(f_1(-1), 1)\} \} &\text {if } \tau_1 = f_1(-1) \\
\inf \{t > f_1(1): B_t \in \{f_2(f_1(1), -1), \ f_2(f_1(1), 1)\} \} &\text {if } \tau_1 = f_1(1)
\end{cases}
$$
Say that $\omega $ is such that $\tau_1(\omega)=f_1(1)$ and $\tau_2(\omega)= f_2(f_1(1), 1)$, then for 
$$f_3(B_{\tau_1 }( \omega), B_{\tau_1 }(\omega) , -1 ) < B_{\tau_2 }(\omega) < f_2(B_{\tau_1 }(\omega), B_{\tau_2}(\omega) ,1)$$
to be true we should have
$$f_3(f_1(1), \ f_2(f_1(1), 1), -1) < f_2(f_1(1), \ 1) < f_3(f_1(1), \ f_2(f_1(1), 1), 1) $$
If there exists $\omega $ such that $X_1(\omega) = f_1(1) $ and $X_2(\omega) = f_2(f_1(1), 1) $ then the last equality follows from 
$$f_3(X_1, X_2, -1) < X_2 < f_2(X_1, X_2, 1)$$
and simarily for the other possible values of $\tau_1 $ and $\tau_2 $.
I suppose we could expand this argument using induction to hold for any $n $.
