Consider the condition that $f:X\to Y$ is continuous iff $f^{-1}(B^o)\subset (f^{-1}(B))^o$.Now the proof can be done as follows,it is obvious that $f^{-1}(B^o)\subset f^{-1}(B)$,this is true for any function.See the diagram below:
Now take interior on both sides $(f^{-1}(B^o))^o\subset (f^{-1}(B))^o$,Now $B^o$ is open in $Y$,so $f^{-1}(B^o)$ is open in $X$ by continuity of $f$ and so $(f^{-1}(B^o))^o=f^{-1}(B^o)$,so $f^{-1}(B^o)\subset (f^{-1}(B))^o$.We are done.I think this is the shortest way to prove the $(\implies)$ part.
The argument is correct, and could be summarised as follows: let $B \subseteq Y$.
Then $$B^\circ \subseteq B$$ (by definition of interior), and as $f^{-1}$ preserves inclusion:
$$f^{-1}[B^\circ] \subseteq f^{-1}[B]$$ and this implies by a standard property of interior (monotonicity), and because inverse images of open sets are open:
$$f^{-1}[B^\circ] = f^{-1}[B^\circ]^\circ \subseteq f^{-1}[B]^\circ$$
which shows the inclusion.
Conversely, if this identity holds for all $B$, and $O \subseteq Y$ is open (so $O^\circ =O$) we have that
$$f^{-1}[O] = f^{-1}[O^\circ] \subseteq f^{-1}[O]^\circ \subseteq f^{-1}[O]$$
which implies that $f^{-1}[O]^\circ = f^{-1}[O]$ and $f^{-1}[O]$ is open, and so $f$ is continuous.