($\implies$) direction was easy enough, but I am confused on $(⟸)$. So we have that If $A \subset R^n$, for every open set $U \in R^m$ there is some open set $V \in R^m$ such that $f^{-1}(U) = V \cap A$. This was my idea:
Let $x \in A$. We have that $f(x) \in R^m$. Let $\epsilon > 0$ be fixed. It is clear that $B_{\epsilon}(f(x))$ is open and since $B_{\epsilon}(f(x)) \subset R^m$, by hypothesis we have that there exists some open set $V \subset R^n$ such that $f^{-1}(B_{\epsilon}(f(x))) = V \cap A.$ This is where I get stuck. If $A = R^n$ this is quite easy to prove because $V$ is open, but since $V \cap A$ isnt necessarily open, im not sure what to do. From the definition of continuity, I want to find some $\delta > 0$ such that for all $a \in A$ satisfying $ 0 < | x - a | < \delta$ that $ | f(x) - f(a) | < \epsilon$. I was thinking that we could do something with the fact that $f^{-1}(R^m) = (\cup_{a \in A} V_{f(a)}) \cap A$. But im also getting no where with this.
Any tips would be greatly appreciated
Edit: Ok so I realised this can be quite easily done (i think) using the idea of a subspace of a topology, but I am still trying to do it with epsilon and delta