We define a sequence of (discrete) stopping times
$$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$
It is not difficult to see that $\tau_j$ is indeed a stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Since the Brownian motion has continuous paths, this implies $B(\tau) = \lim_{j \to \infty} B(\tau_j)$.
Let $\xi,\eta \in \mathbb{R}$. Then, by the dominated convergence theorem,
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \lim_{j \to \infty} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau_j+t)-B(\tau_j))} \cdot e^{\imath \, \eta B(\tau_j)} \bigg) \\ &= \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \xi (B(k \cdot 2^{-j} +t)-B(k \cdot 2^{-j}))} \cdot e^{\imath \, \eta B(k \cdot 2^{-j})} \cdot 1_{\{\tau_j = k \cdot 2^{-j}\}} \bigg) \end{align*}$$
where we used in the last step that $\tau_j$ is a discrete stopping time. By assumption, $B(k \cdot 2^{-j}+t)-B(k \cdot 2^{-j})$ and $B(k \cdot 2^{-j}) \cdot 1_{\{\tau_j=k2^{-j}\}}$ are independent. Therefore, we obtain
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \eta B(k 2^{-j})} \cdot 1_{\{\tau_j=k \cdot 2^{-j}\}} \bigg) \\ &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg). \end{align*}$$
(In the second step we used again dominated convergence, similar to the above calculation.) If we choose $\eta = 0$, then we get
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \bigg) = \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg);$$
hence,
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg)= \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg)$$
i.e. $B(\tau+t)-B(\tau)$ and $B(\tau)$ are independent. Therefore, the strong Markov property gives
$$\begin{align*} \mathbb{E}\bigg(1_F e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid \mathcal{F}_{\tau} \bigg] \bigg) \\ &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid B_{\tau} \bigg] \bigg)\\ &= \mathbb{P}(F) \cdot \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \end{align*}$$
for any $F \in \mathcal{F}_{\tau}$. Consequently, $B(\tau+t)-B(\tau)$ is independent of $\mathcal{F}_{\tau}$.
A very similar calculation shows that
$$\mathbb{E} \left( \exp \left( \imath \sum_{j=1}^n \xi_j \cdot (B(\tau+t_j)-B(\tau+t_{j-1})) \right) \right) = \prod_{j=1}^n \mathbb{E}e^{\imath \, \xi_j B(t_j-t_{j-1})}$$
for any $\xi_j \in \mathbb{R}$, $0 \leq t_0 < \ldots \leq t_n$. This means that $(B(\tau+t_j)-B(\tau+t_{j-1}))_{j=1,\ldots,n}$ are independent normal distributed random variables.
Approach I (via characteristic functions): The identity $$E(1_A g(B_{t_1},\ldots,B_{t_n})) = P(A) E(g(B_{t_1},\ldots,B_{t_n})) \tag{1}$$ can be easily extended to complex-valued continuous functions (just write $g= \text{Re} g + i \, \text{Im g}$ and apply $(1)$ separately to the real and imaginary part of $g$). Choosing $$g(x_1,\ldots,x_n) := \exp \left( i \sum_{j=1}^n \xi_j x_j \right)$$ for some fixed $\xi=(\xi_1,\ldots,\xi_n) \in \mathbb{R}^n$ we find that $$E \left( 1_A \exp \left[ i \sum_{j=1}^n \xi_j B_{t_j} \right] \right) = P(A) E \exp \left( i \sum_{j=1}^n \xi_j B_{t_j} \right)$$ for all $A \in \mathcal{F}_{0+}$ and $\xi \in \mathbb{R}^n$. This implies by Kac's lemma that $\mathcal{F}_{0+}$ and $\sigma(B_{t_1},\ldots,B_{t_n})$ are independent, see the lemma here for details.
Approach II (via monotone class argument): For any open set $U \subseteq \mathbb{R}^n$ there exists a sequence of continuous bounded functions $(g_k)_{k \in \mathbb{N}}$ such that $g_k \uparrow 1_U$. Using $(1)$ and applying the monotone convergence theorem, it follows that \begin{align*} E(1_A 1_U(B_{t_1},\ldots,B_{t_n})) &= \sup_{k \geq 1} E(1_A g_k(B_{t_1},\ldots,B_{t_n})) \\ &= P(A) \sup_{k \geq 1} E(g_k(B_{t_1},\ldots,B_{t_n})) \\ &= P(A) E(1_U(B_{t_1},\ldots,B_{t_n})) \end{align*} for $A \in \mathcal{F}_{0+}$. Since the open sets are a generator of the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$, an application of the monotone class theorem yields that $$E(1_A 1_F (B_{t_1},\ldots,B_{t_n})) = P(A) E(1_F(B_{t_1},\ldots,B_{t_n})), \qquad A \in \mathcal{F}_{0+}$$ for any $F \in \mathcal{B}(\mathbb{R}^n)$. Hence, $\mathcal{F}_{0+}$ and $\sigma(B_{t_1},\ldots,B_{t_n})$ are independent.
Best Answer
Do you refer to corollary 3.19 in this book? I believe that you miss-understand the content of it. What he states in corollary 3.19 is in fact exactly what you're are asking for. Corollary 3.19 states essentially that $P_x( T_r<\infty)=1$ for all $x\in \mathbb{R}^2$. The definition of $T_r$ is
$$T_r:=\inf\{t>0 :| B(t)|=r\}$$ That is; No matter what $x$ we start the brownian motion, it returns to the circle of radius $r$ around 0!
EDIT: Looking at your question again, i realize that you may ask because you are strictly using the form of SMP given in theorem 2.16 stating that the displaced process $B_{T+t}-B_{t}$ is again a brownian motion started at $0$. However I believe that the book refers to the form given in remark 2.17 with $$f(F)=1\{ \exists t>0 : F(t)\in B(x,\varepsilon)\} $$ where the "1" notation means the indicator function. Using the tower property this gives
$P( \exists t>0 : B(t+t_1)\in B(x,\varepsilon)\}) =\mathbb{E}f(B_{t+t_1})\stackrel{2.17}{=}\mathbb{E}f(B_{t})\stackrel{3.19}{=} 1$