This is theorem 1.13:
For every constant $c < \sqrt{2}$, almost surely, for every $\epsilon > 0$ there exist $0 < h < \epsilon$ and $t \in [0,1-h]$ with
$$|B(t+h) - B(t)| \geq c \sqrt{h \log(1/h)} $$
Proof:
Let $c < \sqrt{2}$ and define, for integers $k,n \geq 0 $ the events $$A_{k,n} =\bigg\{B((k+1)e^{-n})-B(ke^{-n}) > c c\sqrt{n}e^{-n/2} \bigg\}$$ then, using lemma 12.9, for any $k \geq 0$, $$\mathbb{P}(A_{k,n}) = \mathbb{P}\{B(e^{-n}) > c\sqrt{n}e^{-n/2}\} = \mathbb{P} \{B(1) > c \sqrt{n} \} \geq \frac{c\sqrt{n}}{c^2n+1} \frac{1}{\sqrt{2\pi}} e^{-c^2n/2}$$
By our assumptions on $c$, we have $e^n \mathbb{P}(A_{k,n}) \to \infty$ as $n \uparrow \infty $. Therefore using $1-x \leq e^{-x}$ for all $x$ $$ \mathbb{P} \bigg( \bigcap_{k=0}^{\lfloor e^{n}-1 \rfloor} A_{k,n}^c\bigg) = (1-\mathbb{P}(A_{0,n}))^{\lfloor e^n \rfloor} \leq \exp(-e^n \mathbb{P}(A_0,n)) \to 0 \text{ as } n \to \infty$$
I understand everything until now, the only part I don't understand is the last one:
By considering $h = e^{-n}$ one can now see that, for any $\epsilon > 0$, $$\mathbb{P} \bigg\{|B(t+h)-B(t)| \leq c \sqrt{h\log(1/h)} \forall h \in (0,\epsilon),t \in [0,1-h] \bigg\} = 0 $$ And that's all they wrote. I understand that this last equality implies the result, what I don't understand is how they got this result. First of all $h = e^{-n}$ so I guess it is an abuse of notation to use $h$ inside the event. Also, by replacing $h$ in the definition of $A_{k,n}$ one can get $|B(kh+h) - B(kh)| > c \sqrt{h\log(1/h)}$ but I don't see how this implies the last equality.
Note that $$\{|B(t+h)-B(t)|\leq c\sqrt{h log(\frac{1}{h})} \ \forall h \in (0,\epsilon), t\in[0,1-h]\} \\ \subset \{B(t+h)-B(t)\leq c\sqrt{h log(\frac{1}{h})} \ \forall h \in (0,\epsilon), t\in[0,1-h]\} \\ \subset \{B((k+1)2^{-n})-B(k2^{-n})\leq c\sqrt{n}e^{-\frac{n}{2}} \ \forall k=0,1,\dots ⌊^n−1⌋\} $$ for any $n$ so that $e^{-n}<\epsilon$. But $$ \{B((k+1)2^{-n})-B(k2^{-n})\leq c\sqrt{n}e^{-\frac{n}{2}} \ \forall k=0,1,\dots ⌊^n−1⌋\}=\bigcap_{k=0}^{⌊^n−1⌋}A_{k,n}^c $$ so $$ \mathbb{P}(\{|B(t+h)-B(t)|\leq c\sqrt{h log(\frac{1}{h})} \ \forall h \in (0,\epsilon), t\in[0,1-h]\}) \leq \mathbb{P}(\bigcap_{k=0}^{⌊^n−1⌋}A_{k,n}^c) $$ for sufficiently large $n$. Taking $n \to \infty$ yields the result.