Exercise 1.13 of chapter 1 of Revuz and Yor's This is the exercise 1.13 of chapter 1 of Revuz and Yor's.

  
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*Let $B$ be the standard linear BM. Prove that $\varlimsup_{t\to\infty}(B_t/\sqrt{t})$ is a.s. $>0$ (it is in fact equal to $+\infty$ as will be seen in Chap. 2)
  
*Prove that $B$ is recurrent, namely: for any real $x$, the set $\{t: B_t=x\}$ is unbounded.
  
*Prove that the Brownian paths are a.s. nowhere locally Holder continuous of order $\alpha$ if $\alpha>\frac{1}{2}$.
  [Hint: Use the invariance properties of Proposition (1.10).]
  

What I tried:
For 1. $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1-P(\varlimsup_{t\to\infty}B_t/\sqrt{t}\leq  0)$ and fix $\varepsilon>0$ there exist large enough $t_0$ s.t. $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}\leq 0)\leq P(B_{t_0}/\sqrt{t_0}<\varepsilon)$<1/2, then $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)>1/2$. By Hewitt-Savage zero-one law, $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1$. But I am not very familier with HS 0-1 law, is there some other method to solve this problem? (also not using law of iterated logarithm)
For 2. By 1, $P(\varliminf_{t\to\infty}B_t/\sqrt{t}<0)=P(\varlimsup_{t\to\infty}(-B_t)/\sqrt{t}>0)=P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1$, so $x=0$ is recurrent. But how to prove it when $x\neq0$?
For 3. It is sufficient to prove $P\left(\sup_{0\leq s,t\leq 1}\frac{|B_t-B_s|}{\sqrt{|t-s|}}=+\infty\right)=1$. It can be proved by the independent increment property and scaling property of Brownian Motion.

Re 1., note that the random variable $X=\limsup\limits_{t\to\infty}B_t/\sqrt{t}$ is asymptotic hence, by Kolmogorov zero-one law, if $P(X\gt0)\lt1$, then $P(X\gt0)=0$, that is, $X\leqslant0$ almost surely. If this holds, then, by symmetry, $\liminf\limits_{t\to\infty}B_t/\sqrt{t}\geqslant0$ almost surely, thus, $B_t/\sqrt{t}\to0$ almost surely. 
In particular, this would imply that $B_t/\sqrt{t}\to0$ in distribution, which is absurd since every $B_t/\sqrt{t}$ is standard normal. Hence $X\gt0$ almost surely.
