sIt is the space $W^{1,1}[0, 1]$. Note initially that $W^{1,1}[0, 1]$ is a subspace of $BV$ moreover in there the norms are equivalent. In fact if $f\in W^{1,1}[0, 1]$, for its continuous representative
f(x)=f(0)+\int_{0}^{x}f'(t)dt\tag{1} Then |f(x)|\leq|f(0)|+|\int_{0}^{x}f'(t)dt|\leq |f(0)|+\int_{0}^{1}|f'(t)|dt.
But V_f[0,1]=\int_{0}^{1}|f'(t)|dt. So $||f||_L^1\leq||f||_{\infty}\leq||f||_{BV}$ since ||f'||_L^1\leq||f||_{BV} then
$||f||_{W^{1,1}}\leq 2||f||_{BV}.$ Using $(1)$ again we get f(0)=-f(x)+\int_{0}^{x}f'(t)dt
then |f(0)|\leq |f(x)|+\int_{0}^{1}|f'(t)|dt,
integrating from $0$ to $1$ $|f(0)|\leq ||f||_{W^{1,1}}$ and analogously $||f||_{BV}\leq 2||f||_{W^{1,1}}$ ergo both norms are equivalent.
If $f_n \to f$ in $BV$ with $(f_n)\subset C^{1}$ then $(f_n)$ is a Cawchy sequence in $W^{1,1}$, since it is a complete space $f_n\to g$ in $W^{1,1}$.
Then $f_n\to g$ in $BV$ because the norms are equivalents and by the unity of the limit $g=f\in BV$.
This proves that the closure of ${C^1}$ in the $BV$ norm is a subspace of $W^{1,1}$.
The other inclusion is given because any function $f\in W^{1,1}$ can be approximated by a sequence $f_n\to f $ in $W^{1,1}$ what is the same $f_n\to f $ in $BV$ $\blacksquare$
PS: That is my answer it may contains some English problems but it is mathematically correct! And all details are included! It is of course inspired on @Jonas answer! And he deserved to gain the bounty, and I thanks to him! But how he was not so sure ( he wrote I think it is W1,1) I had to put a complete answer. My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer.