The fact that $F_n$ is closed is quite elementary.Take any partition $\{x_0,x_1,...,x_N\}$ of $[0,1]$. If $f_k \in F_n$ for all $k$ then $\sum |f_k(x_i)-f_k(x_{i-1})| \leq n$ for all $k$. If $f_k \to f$ in the sup norm then it converges point-wise so we get $\sum |f(x_i)-f(x_{i-1})| \leq n$. Take sup over all partitions to get $V(f) \leq n$.


For the first question, I will prove that $F_1$ is closed. So the proof will work for any $F_n$.

Let $(f_n)_{n\in \mathbb{N}}$ be a convergent sequnce in $F_1$, say $f_n\rightarrow f_0$ for some $f_0\in BV[a,b]\cap C[a,b]$. We wish to show that for any partition $P$ we have $V(f_0,P)<1$. So let $P=(a=t_1,t_2,…,t_k=b)$ be a partition for $[a,b]$. Let $\epsilon _n =\dfrac{1}{n}$ for all $n$. Then for any natural number $n$, there exists a natural number $N_n$ such that $sup_{x\in [a,b]} |f_{N_n}(x)-f_0(x) | < \dfrac{\epsilon _n}{2k}$. For all $n$, we have $V(f_0,P)=\displaystyle \sum _{i=1}^{k-1} |f_0(t_{i+1})-f_0(t_i)| \leq \displaystyle \sum _ {i=1}^{k-1} |f_0(t _{i+1})-f_{N_n}(t _{i+1} )|+|f_{N_n}(t_{i+1})-f_{N_n}(t_i)|+| f_{N_n}(t_i)-f_0(t_i)|$

$\leq 1+ 2k \dfrac{\epsilon _n}{2k}=1+\epsilon _n $. If we take the limit as $n\rightarrow \infty$, we get $V(f_0,P)\leq 1$. Hence, $f_0\in F_1$.

I assumed completeness of $C([a,b])$ is known.