I quote Kuo (2006)
Keeping in mind the Hewitt-Savage $0-1$ law for exchangeable events, that is:
If $E$ is an exchangeable event for an independent identically distributed sequence, than $\mathbb{P}(E)$ is either $0$ or $1$.
I am trying to understand the proof of the following proposition:
Proposition. Let $\{B(t):t\ge0\}$ be a standard Brownian motion. Almost surely, $$\limsup\limits_{n\to\infty}\frac{B(n)}{\sqrt{n}}=+\infty\hspace{0.5cm}\text{ and }\hspace{0.5cm}\liminf_{n\to\infty}\frac{B(n)}{\sqrt{n}}=-\infty$$
Herebelow, I write the proof and - in $\color{red}{\text{ bold red }}$ - my doubts/observations:
Proof of proposition. By Fatou's lemma: $$\mathbb{P}\{B(n)>c\sqrt{n}\text{ infinitely often}\}=\mathbb{P}\{\limsup_{n\to\infty}(B(n)>c\sqrt{n})\}\ge\limsup_{n\to\infty}\mathbb{P}\{B(n)>c\sqrt{n}\}$$ By the scaling property, the expression in $\limsup$ equals $\mathbb{P}\{B(1)>c\}$, which is positive. $\color{red}{\text{($1.$ Why is it positive? Couldn't it be $0$?)}}$
Let $X_n=B(n)-B(n-1)$ and note that $$\{B_n>c\sqrt{n}\text{ infinitely often}\}=\{\sum_{j=1}^n X_j>c\sqrt{n}\text{ infinitely often}\}$$ is an exchangeable event.
$\color{red}{\text{($2.$ Why is that an exchangeable event (by def.))?}}$
Hence, the Hewitt-Savage $0-1$ law gives that, with probability one, $B(n)>c\sqrt{n}$.
$\color{red}{\text{($3.$ As stated above, Hewitt-Savage $0-1$ law establishes that probability associated to}}$ $\color{red}{\text{ an exchangeable event is either $0$ or $1$. So, why could we say with certainty that }}$ $\color{red}{\text{$\mathbb{P}(B(n)>c\sqrt{n}$ infinitely often)$=1$ and not that $\mathbb{P}(B(n)>c\sqrt{n}$ infinitely often)$=0$?)}}$
Taking the intersection over all positive integers $c$ gives the first part of the proposition and the second part is proven analogously.
$\color{red}{\text{($4.$What is exactly meant here? One shall take intersection of which events? How can}}$ $\color{red}{\text{this allow to pass from}}$ $\color{red}{\text{$\mathbb{P}(B(n)>c\sqrt{n}\text{ infinitely often})=1$ to $\limsup\limits_{n\to\infty}\frac{B(n)}{\sqrt{n}}=+\infty$?)}}$
Note that $\{\limsup B_n/\sqrt{n}\ge c\}\supseteq \{B_n/\sqrt{n}> c\text{ i.o.}\}\equiv A$ and $$ \{\limsup B_n/\sqrt{n}=+\infty\}=\bigcap_{c\in \mathbb{N}}\{\limsup B_n/\sqrt{n}\ge c\}. $$ If the probability of each event in the intersection is $1$, then the probability of the intersection of these events is also $1$. Since $\mathsf{P}(A)\in \{0,1\}$ ($\because A$ is an exchangeable event) and $\mathsf{P}(A)\ge \mathsf{P}(B_1>c)>0$, the result follows.
Exchangeable event. An event $E\in \sigma(X_1,X_2,\ldots)$ is exchangeable iff there exists $B\in \mathcal{B}(\mathbb{R}^{\infty})$ s.t. for each $n\ge 1$ and any permutation $\pi_n$ of $\{1,\ldots, n\}$, $$ E=\{(X_1,\ldots,X_n,X_{n+1}\ldots)\in B\}=\{(X_{\pi_n(1)},\ldots,X_{\pi_n(n)},X_{n+1},\ldots)\in B\}. $$
In your case $X_n:=B_{n}-B_{n-1}$, $n\ge 1$ with $B_0\equiv 0$, and $E=\{\sum_{i= 1}^n X_i> c\sqrt{n}\text{ i.o.}\}$.