This is my answer:
1) $\to$ 2).
If $f^{-1}(V)=\emptyset$ the answer is trivial.
Let $a\in f^{-1}(V)$. Particularly $a\in A$. As $f$ continuous in $A$ then $\forall \varepsilon>0 \exists\delta>0$ such that, if $||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon$. i.e:
if $x\in \mathbb{B}_{\delta}(a) \implies f(x)\in \mathbb{B}_{\varepsilon}(f(a))$
$\implies f(\mathbb{B}_{\delta}(a))\subset \mathbb{B}_{\varepsilon}(f(a))$
$\implies \mathbb{B}_{\delta}(a)\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a)))$
$\therefore f^{-1}(V)$ is open.
2) $\to$ 1).
Let $\varepsilon>0$ and $a\in A$. $V$ is open then exists $\mathbb{B}_{\varepsilon}(f(a)))$ open. Now $a\in f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ is open. i.e:
$\exists \delta$ such that $\mathbb{B}_{\delta}(a))\subset f^{-1}(\mathbb{B}_{\varepsilon}(f(a))))$ (you can prove that this is true)
$\implies f(\mathbb{B}_{\delta}(a)))\subset \mathbb{B}_{\varepsilon}(f(a)))$
$||x-a||<\delta \implies ||f(x)-f(a)||<\varepsilon $
$\therefore f$ is continuous.