Now I have some trouble understanding what $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\}$ means.
It is nothing but $\mathbb{P}_{x}\{y\in B[0,1]\}|_{x=B(\varepsilon)}$.
I read this notation as follows: the event consists of all $\omega$ such that $y\in B([0,1],\omega)$ where $B$ is a Brownian motion starting at $B(\varepsilon,\omega)$.
No. The event precisely consists of all $\omega$ such that $y \in B([0,1],\omega)$. For instance, if $B$ is defined canonically (i.e. with $\Omega = C([0,1];\mathbb R^2)$ and $B(\cdot,\omega) =\omega(\cdot)$), then this event is nothing else but the set of functions attaining value $y$ at some point. This has nothing to do with the probability $\mathbb P_{x}$ (actually, probabilities, since there are many of them, and they are mutually singular). Naturally, since the value $\mathbb P_x(A)$ is zero for almost all $x$ and since the law of $B(\varepsilon)$ is absolutely continuous, then $\mathbb P_{B(\varepsilon)}(A) = 0$ almost surely.
Actually, the authors are indeed a bit sloppy here: as I've already mentioned, there are many probability measures, so "almost surely" may a priori sound pretty ambiguous. A diligent way of writing this is "for any $x$, $\mathbb P_{B(\varepsilon)}(A) = 0 \ $ $\ \mathbb P_x$-almost surely". And they are using precisely this when they write further (thanks to the Markov property and homogeneity) that
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb E_x \mathbb P_{B(\varepsilon)} (y \in B ([0,1-\varepsilon]))=0.
$$
Some elaboration of the last formula:
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) = y\}) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + B(\varepsilon) = y\})
$$
Thanks to the ($\mathbb P_x$-)independence of $B(t) - B(\varepsilon)$ and $B(\varepsilon)$, this is equal to
$$
\mathbb E_x \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + z = y\})\big|_{z = B(\varepsilon)}.
$$
Now using the fact that (under $\mathbb P_x$) $B(t) - B(\varepsilon)$ is, up to a time shift by $\varepsilon$, a Brownian motion starting from $0$, we can further rewrite this as
$$\mathbb E_x \mathbb P_0 (\{\exists s\in [0,1-\varepsilon]: B(s) + z = y\})\big|_{z = B(\varepsilon)} = \mathbb E_x \mathbb P_z (\{\exists s\in [0,1-\varepsilon]: B(s) = y\})\big|_{z = B(\varepsilon)}.
$$
About the small increments $B(t+h)-B(t)$ of BM(Brownian motion), there are two classes of results:
(i) LIL(Law of Iterated Logarithm) for fixed start point $t$;
(ii) Levy's modulus of continuity for moving start points.
(i) For fixed start point--LIL :
\begin{gather*}
\varlimsup_{\delta\downarrow0}\frac{|B(t+\delta)-B(t)|}{ \sqrt{2\delta \log\log\delta^{-1}}}=1.\quad \text{a.s.}\quad \forall t>0. \\
\varlimsup_{\delta\downarrow0} \frac{\sup_{0< h\le \delta} |B(t+h)-B(t)|}{\sqrt{2\delta\log\log \delta^{-1}}}=1.\quad \text{a.s.} \quad \forall t>0.
\end{gather*}
(ii) For moving start points--Levy's modulus:
\begin{gather*}
\varlimsup_{\delta\downarrow0}\frac{\sup_{0< t\le 1}|B(t+\delta)-B(t)|}{ \sqrt{2\delta \log\delta^{-1}}}=1.\quad \text{a.s.}\\
\varlimsup_{\delta\downarrow0}\frac{\sup_{0< t\le 1} \sup_{0< h\le \delta} |B(t+h)-B(t)|}{\sqrt{2\delta\log\delta^{-1}}}=1.\quad \text{a.s.}\\
\varlimsup_{\delta\downarrow0}\frac{\sup_{0<s, t\le 1, |s-t|<\delta} |B(t)-B(s)|}{\sqrt{2\delta\log\delta^{-1}}}=1.\quad \text{a.s.}
\end{gather*}
Remark: in the book M. Csörgo and P.Révész, Strong Approximations in Probability and Statics, Academic Press, 1981. $\S1.2$ p.29--., there are more results about "how big are the increments of a Wiener process".
Best Answer
The Brownian motion $B$ induces a measure $\mu$ ("Wiener measure") on the measurable space $(C[0,1],\mathcal C)$ via the map $\Phi:\Omega\to C[0,1]$ defined by $\Phi(\omega) := (B(t,\omega))_{0\le t\le 1}$. The event $$ G:=\left\{x\in C[0,1]: \limsup_{h\to 0}\sup_{0\le t\le 1-h}{|x(t+h)-x(t)|\over C\sqrt{h\log(1/h)}}>a\right\} $$ is an element of $\mathcal C$ and $$ \mu(G)=\Bbb P(\Phi^{-1}(G))=\mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|B(t+h)-B(t)|}{C\sqrt{h\log(1/h)}}>a\right) = 0. $$ Because $\tilde B$ is also Brownian motion, it induces the same measure $\mu$, and so $$ \mathbb{P}\left(\limsup_{h\rightarrow 0}\sup_{0\leq t\leq 1-h}\frac{|\tilde B(t+h)-\tilde B(t)|}{C\sqrt{h\log(1/h)}}>a\right) = \mu(G) =0, $$ as well.