Great question: I really had no idea how it was going to go. The answer is that $f[A^\circ]\supseteq f[A]^\circ$ is neither sufficient nor necessary for $f$ to be continuous.
Counterexample to Sufficiency
Generalizing from an example given in the comments, just send $X$ discontinuously to a set with empty interior. Then every subset of $X$ goes to a set with empty interior as well, and the condition holds trivially. It's not hard to make the function discontinuous: as long as $X$ isn't discrete and its image intersects two different open sets (e.g. is at least two points of a Hausdorff space), we can do it.
Counterexample to Necessity
Let $f:(X,\tau)\rightarrow (X,\sigma)$ where $X=\{a,b,c\}, \tau=\{X,\{a\},\{b,c\}\}, \sigma=\{X,\{a\},\{b\},\{a,b\}\}.$ Set $f(b)=f(c)=b, f(a)=a$. Then $f$ is continuous, since $f^{-1}(\{a,b\})=X,$ $f^{-1}(\{b\})=\{b,c\}$, and $f^{-1}(\{a\})=\{a\}$. But on the other hand $\{a,b\}$ has empty interior and is mapped to an open set $\{a,b\}$.