Let $A_t:=\int_0^tf(B_s)ds,\quad t\geq0$
with $f$ continuous and $B$ standard Brownian motion.
What is the correct argument that $A$ is of finite variation? Because it can be written as an integral?
And why is $\int_0^t f(B_s)dB_s$ a local martingale? Because $B$ is a local martingale and $f$ is continous?
Since $B_s$ has continuous paths, $f(B_s)$ has continuous paths and hence, $A_t$ has $C^1$ paths (that's the fundamental theorem of calculus). However, any $C^1$ process has locally bounded variation (some authors omit this word) since any $C^1$-function has locally bounded variation.
As for why $\int_0^t f(B_s)\textrm{d}Bs$ is a local martingale, that simply follows from the construction of the Itô integral.