Let me give you some extra tools to use, here. Some details are left to you to verify.
Definitions: A subset $A$ of a topological space $X$ is said to be nowhere-dense (in $X$) if for every non-empty open subset $U$ of $X$ there is a non-empty open subset $V$ of $X$ such that $V\subseteq U$ and $V\cap A=\emptyset$. (You should be able to prove that this is equivalent to your definition of nowhere-dense.) A topological space $X$ is said to be a Baire space if every countable union of nowhere-dense subsets of $X$ has empty interior.
Lemma 1: $X$ is a Baire space if and only if every countable intersection of open dense subsets of $X$ is dense in $X$.
Proof: Suppose $X$ is a Baire space and $\{G_n\}_{n=1}^\infty$ is a countable collection of open dense sets, so each $X\setminus G_n$ is nowhere dense. (Why?) Thus, for any non-empty open set $U$, we have that $$U\nsubseteq\bigcup_{n=1}^\infty (X\setminus G_n)=X\setminus\left(\bigcap_{n=1}^\infty G_n\right),$$ so there must be some point of $\bigcap_{n=1}^\infty G_n$ in $U$. Thus, $\bigcap_{n=1}^\infty G_n$ is dense.
Suppose countable unions of open dense sets are dense, and let $A_n$ be nowhere dense for each $n$, so each $X\setminus\overline{A_n}$ is open and dense. (Why?) Take any non-empty open $U$, so that $U$ intersects $$\bigcap_{n=1}^\infty\left(X\setminus\overline{A_n}\right)=X\setminus\left(\bigcup_{n=1}^\infty\overline{A_n}\right)\subseteq X\setminus\left(\bigcup_{n=1}^\infty A_n\right),$$ so cannot be a subset of $\bigcup_{n=1}^\infty A_n$. Thus, $X$ is a Baire space. $\Box$
Lemma 2: $X$ is a Baire space if and only if every countable union of closed nowhere-dense subsets of $X$ has empty interior. (I leave the proof to you. Lemma 1 and DeMorgan's Laws should help.)
Lemma 3: If $X$ is a non-empty complete metric space with metric $d:X\times X\to\Bbb R$, and $F$ is a non-empty closed subset of $X$, then $F$ is a complete metric space with metric $d_F$ defined by $d_F(x,y)=d(x,y)$ for all $x,y\in F$. (I leave the proof of this to you.)
Proposition: Assuming BCT holds, every non-empty complete metric space is a Baire space.
Proof: Suppose by way of contradiction that $X$ is a non-empty complete metric space, but isn't a Baire space. Then by Lemma 2, there is some countable collection $\{A_n\}_{n=1}^\infty$ of closed nowhere-dense subsets of $X$ whose union doesn't have empty interior--meaning there is some non-empty open $U$ such that $$U\subseteq\bigcup_{n=1}^\infty A_n.$$ Now, in particular, we can take a non-empty open set $V$ such that $\overline V\subseteq U.$ (Why?) Then $$\overline V\subseteq\bigcup_{n=1}^\infty A_n,$$ so $$\overline V=\overline V\cap\left(\bigcup_{n=1}^\infty A_n\right)=\bigcup_{n=1}^\infty(\overline V\cap A_n).\tag{#}$$ But $\overline V$ is a non-empty complete metric space by Lemma 3, so by BCT, $\overline V$ is not a countable union of closed nowhere-dense subsets of $\overline V$. But each $\overline V\cap A_n$ is closed and nowhere-dense in $\overline V$ (why?), so $(\#)$ gives us the desired contradiction. $\Box$
Best Answer
It only lasts to see that $\displaystyle\bigcap_{n\in\mathbb N} A_n \subset \mathbb B$.
Fix $f\in\displaystyle\bigcap_{n\in\mathbb N} A_n$ and we will prove below that $f\in\mathbb B$.
Fix $t\in[0,1]$. For each $n\in \mathbb N$, there exists $h_n$ such that $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right| > n. $$ Therefore,
Now observe that
Suppose the contrary. Then, there exists $\delta>0$ such that $|h_n|\geq \delta$ for any $n\in\mathbb N$. Then $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right|\leq \dfrac{|f(t+h_n)-f(t)|}{\delta},\ \forall n \in \mathbb N, $$ and since $f$ is continuous in $[0,1]$, it is bounded by some $M>0$, and so $$ \left|\dfrac{f(t+h_n)-f(t)}{h_n}\right|\leq \dfrac{2M}{\delta},\ \forall n \in \mathbb N, $$ but this contradicts the fact that the limit is infinite, and the claim is proved
Fix the subsequence $(h_{n_m})_{m\in\mathbb N}$ given by the Claim. We have that $h_{n_m}\to 0$ and $$ \lim_{m\to +\infty}\left|\dfrac{f(t+h_{n_m})-f(t)}{h_{n_m}}\right|=+\infty, $$ so using this classical equivalence on limits (from real analysis):
it follows that the limit $\displaystyle\lim_{h\to0}\left|\dfrac{f(t+h)-f(t)}{h}\right|$ doesn't exist, and consequently, $\displaystyle\lim_{h\to0}\dfrac{f(t+h)-f(t)}{h}$ doesn't exist as well, so $f$ is not differentiable at $t$. Since $t\in[0,1]$ was fixed arbitrarily, it follows that $f$ is nowhere differentiable and so $f\in\mathbb B$.