Give an example of a Banach space $V$, a normed space $W$, a bounded linear surjective map $T: V \to W$ and an open subset $G \subseteq V$ such that $T(G)$ is not open in $W$.
Attempt: Consider $V= (C([0,1], \Vert \cdot \Vert_\infty), W= (C([0,1], \Vert \cdot \Vert_1)$ and $T: V \to V: f \mapsto f$. Clearly $T$ is a linear surjection with $$\Vert Tf \Vert_1 = \int_0^1 |f| \le \int_0^1 \Vert f \Vert_\infty = \Vert f \Vert_\infty$$
so $\Vert T \Vert \leq 1$ and $T$ is bounded. Moreover, we have $\Vert f \Vert_1 \leq \Vert f \Vert_\infty$.
We now show that $G= B_\infty(0,1)$ is not open for $\Vert \cdot \Vert_1$. Indeed, suppose to the contrary that $0$ is a $\Vert \cdot \Vert_1$-interior point of $G$. Then there is $\epsilon > 0$ such that
$$B_1(0, \epsilon) \subseteq G = B_\infty(0,1)$$
Thus, for $f \in C([0,1])\setminus \{0\}$ we have $$\Vert \frac{\epsilon}{2 \Vert f \Vert_1} f \Vert_\infty \leq 1$$
I.e. $\Vert f \Vert_\infty \leq \frac{2}{\epsilon} \Vert f \Vert_1$ for $f \in C([0,1])$. But then the norms $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_\infty$ are equivalent, which implies that $W$ is Banach. This is a contradiction.
Question: Is my attempt correct?
Yes, your attempt is correct.
The equivalence of the norms requires both inequalities, so it would be good to mention that.
I think it could also be improved by explaining why $W$ is not a Banach space.