Theorem: Let $D=B(0,r)=\{y:|y|<r\}$. $B_t$ will return to $D$ i.o. in $d\leq 2$ in a $d$-dimensional Brownian motion. Then for any $x$
$$ \begin{align} P_x \left( \int_0^{\infty} 1_D(B_t)dt = \infty \right) &= 1 \text{ for $d\leq 2$ } \\ \end{align} $$
I have a couple of questions about the proof
Proof: Let $T_0 = 0$ and $G = B(0,2r)$. For $k\geq 1$, let
$$ \begin{align} S_k &= \inf\{t> T_{k-1}:B_t\in D\} \\ T_k &= \inf\{t> S_{k}:B_t\in G\} \\ \end{align} $$
Writing $\tau$ for $T_1$ and using the strong Markov property, we get for $k\geq 1$:
$$P_x\left( \int_{S_k}^{T_k}1_D(B_t)dt\geq s\Bigg|\mathcal{F}_{S_k}\right) = P_{B(S_k)}\left( \int_0^{\tau}1_D(B_t)dt \geq s \right)$$
From the strong Markov property it follows that
$$\int_{S_k}^{T_k}1_D(B_t)dt \;\;\;\;\text{ are i.i.d.}$$
Since these random variables have positive mean, it follows from the strong law that:
$$\int_0^\infty 1_D(B_t)dt\geq \sum_{k=1}^{\infty}\int_{S_k}^{T_k}1_D(B_t)dt = \infty \;\;\text{ a.s.}$$
1) From the definition of $T_k$, isn't $T_k = 0$ a.s.? If we start from a point in $D$, any Brownian path that hits $G$ has to go through $G$? I am confused about this.
2) Why are $\int_{S_k}^{T_k}1_D(B_t)dt$ identically distributed? If we start from a point in $D$, why doesn't the probability distribution of $\int_{S_k}^{T_k}1_D(B_t)dt$ depend on where we start from in $D$?