The weak topology is weaker/coarser than the strong topology, so any weakly open subset of $Y$ is strongly open. Therefore, if $f : X \rightarrow Y$ is strong-to-strong continuous, that is $f^{-1}(\mathcal{U})$ is open for any strongly open $\mathcal{U} \subseteq Y$, then $f^{-1}(\mathcal{U})$ is open for any weakly open $\mathcal{U} \subseteq Y$. That is, $f$ is automatically strong-to-weak continuous.
However, $f$ is not necessarily weak-to-strong continuous. I'll get to a counterexample in a moment, but it's worth pointing out that weak-to-strong continuity implies strong-to-strong continuity. To say $f$ is weak-to-strong continuous, we mean that $f^{-1}(\mathcal{U})$ is weakly open in $X$ for any strongly open $\mathcal{U} \subseteq Y$. But again, $f^{-1}(\mathcal{U})$ being weakly open implies it is also strongly open, so $f$ is strong-to-strong continuous.
As a counterexample, let $X$ be any real infinite-dimensional space, $Y$ be $\mathbb{R}$, and $f(x) = \|x\|$. As I'm sure you're aware, $f$ is strong-to-strong continuous. Consider the inverse image $f^{-1}(-1, 1)$ of the strongly open subset $(-1, 1) \subseteq \mathbb{R}$. It's not difficult to see that $f^{-1}(-1, 1)$ is the open unit ball of $X$.
This set is bounded, and hence not weakly open. It's not too difficult to appreciate geometrically why this is the case ($X$ being infinite-dimensional is necessary here). Basic open sets in the weak topology have the form
$$\mathcal{U} = \lbrace x \in X : f_i(x) \in (a_i, b_i), \forall i = 1, \ldots, n\rbrace$$
for a family of bounded linear functionals $f_1, \ldots, f_n$ and open intervals $(a_1, b_1), \ldots, (a_n, b_n) \subseteq \mathbb{R}$. Pick any $c_i \in (a_i, b_i)$, and the set
$$Z = \lbrace x \in X : f_i(x) = c_i \rbrace \subseteq \mathcal{U}$$
is an affine subspace (a linear subspace that has possibly been translated away from the origin) of finite codimension. Since $X$ is infinite-dimensional, this implies $Z$ is infinite-dimensional, thus not the trivial $\lbrace 0 \rbrace$ space, and in particular, unbounded! So, every basic weakly open set is unbounded, and since every weakly open set contains one of these unbounded set, every weakly open set is unbounded in $X$.
Hope that helps!
In my opinion yes, since you know have a new topology, you should prove again that the new topology
leads to a topological vector space.
Your attempt in the comments sounds correct to me so far.
Of course, one should also show that the multiplication by a scalar is continuous.
Best Answer
We can prove it by double inclusion.
One direction follows from the fact that the weak topology has less open sets than the topology induced by the norm.
For the other direction, we can use the geometric form of Hahn-Banach theorem.