The point of the exercise is to show the necessity of requiring both normed spaces to be Banach for the open mapping theorem (which states that any surjective bounded linear operator $T: X\to Y$ between Banach spaces is open).
Let $Y$ be any Banach space with norm $||\cdot||$, and take $X$ to be the same vector space, but with the norm $||x||_\omega := ||x|| + |\omega(x)|$ for any non-continuous functional $\omega$. Take the map $\operatorname{Id}: X \to Y$. Prove that this is a bounded, bijective map but it is not open.
$\textbf{Bounded and bijective:}$ Bijective is trivial, and bounded follows from $||x|| \leq ||x||_\omega$ for all $x$.
$\textbf{Not open:}$ To show this I thought to take a specific open ball and show it does not get mapped to an open set in $Y$. In other words, a subset $U$ that is open for the $\omega$ - norm but not the regular norm.
Take the open ball $B_\omega(0,1) = \{x \in X \mid ||x||_\omega < 1\}$. I wanted to show this is not open for the regular norm, by showing that there is some $y \in B_\omega(0,1)$ such that for all $\varepsilon > 0$, the intersection $B(y,\varepsilon) \cap B_\omega(0,1)^C \neq \emptyset$ (here $B(y,\varepsilon)$ denotes the open ball w.r.t. the regular norm). I thought maybe to take $y=0$ (since this is the only actual element we know is in that $\omega$ - open ball). I thought maybe that because $\omega$ is non-continuous, then for our choice of $\varepsilon$, we could find a $x \in X$ such that $|\omega(x)| > \varepsilon ||x||$. This would ensure that, by taking $||x|| \geq \frac{1}{1+\varepsilon}$, we have $||x||_\omega \geq 1$ which means it is in $B_\omega(0,1)^C$. However, we do not have $||x|| < \varepsilon$ then.
Any idea how to fix this issue?
You are almost there. What you need is $z$ such that $$ |\omega(z)|>\frac2\varepsilon\|z\|. $$ Now you can take $$ x=\frac\varepsilon{2\|z\|} z. $$ Then $\|x\|<\varepsilon$ and $$ |\omega(x)|=\frac\varepsilon{2\|z\|}\,|\omega(z)|>\frac\varepsilon{2\|z\|}\,\frac{2\|z\|}\varepsilon = 1. $$