$\newcommand\F{\mathcal F}\newcommand\N{\mathbb N}$First, some preliminary remarks:
Your link to Bertoin's book is not very good. Here is a link with a better reference to the book.
It is Blumenthal, not Bulmenthal.
It is not true that "when $B$ is closed $\tau_B=\tau'_B$ is failed on $\partial B$". This kind of failure may occur only at some (not all) points on $\partial B$, which are then called irregular points for $B$.
In Bertoin's book, $(X_t)_{t\ge0}$ is a right-continuous Lévy process, and $\F_t$ is the completion of the sigma-field generated by $(X_s\colon0\le s\le t)$. According to the book, let $T_B:=\tau_B$ and $T'_B:=\tau'_B$.
Now, to your question, "how to use Bulmenthal 0-1 law":
Let $B$ be a closed set. By part (iii) of Corollary 8 on p. 22 in Bertoin's book, $T'_B$ is a stopping time and hence $[T'_B=0]=[T'_B\le0]\in\F_0$, where $[\cdots]$ means the event $\cdots$. So, by the definition of $\F_t$ cited above in Remark 4, we see that $P_x(T'_B=0)$ is $0$ or $1$ for any $x$. Suppose now that a point $x\in\partial B$ is irregular for $B$. Then $P_x(T_B=T'_B)\ne1$, while $P_x(T_B=0)=1$. So, $P_x(T'_B=0)\ne1$. Thus, $P_x(T'_B=0)=0$.
In the above proof we did not explicitly use the Blumenthal 0-1 law, which states that the filtration $(\F_t)$ is right-continuous at $0$, that is, $\F_{0+}=\F_0$. However, in the proof in Bertoin's book of the fact that $T'_B$ is a stopping time he used (not quite explicitly) Proposition 4 on p. 18 in that book, which states that the filtration $(\F_t)$ is right-continuous at all real $t\ge0$, and the proof of Proposition 4 on p. 18 in that book is based on Kolmogorov's 0-1 law. So, this reference in Bertoin's book to the Blumenthal 0-1 law is indeed imprecise and possibly confusing.