I might have come up with the answer myself, I wonder if it is correct.
Let $(Y,||\cdot||_Y)$ be an infinite dimensional Banach space and let $g:Y\to\mathbb{R}$ be an unbounded linear functional. Let $X:=(Y,||\cdot||_g)$ where $||y||_g:=||y||_Y+|g(y)|$ for $y\in Y$. One can easily check that $||\cdot||_g$ defines indeed a norm. Let $T:X\to Y$ be the identity operator. Then $T$ is linear and bijective. We can calculate
$$
\|T\|_{op}=\sup\{||Tx||_Y\,\big|\,x\in Y,||x||_g\leq 1\}=\sup\{||x||_Y\,\big|\,x\in Y,||x||_g\leq 1\}\leq\sup\{||x||_Y+|g(x)|\,\big|\,x\in Y ||x||_g\leq 1\}=\sup\{||x||_g\,\big|\,x\in Y ,||x||_g\leq 1\}=1.
$$
It follows that $T$ is bounded.
We will now give a prove by contradiction. Let us assume $T$ is open. Then $T$ is continuous open bijection, hence $T$ is a homeomorphism. It follows that $T^{-1}$ is continuous, linear hence bounded. However, we calculate that
$$
\|T^{-1}\|_{op}=\sup\{||T^{-1}(x)||_Y\,\big|\,||y||_Y\leq 1\}=\sup\{||y||_Y+|g(y)|\,\big|\,||y||_Y\leq 1\}=1+\sup\{|g(y)|\,\big|\,||y||_Y\}=1+\|g\|_{op}=\infty.
$$
So $T^{-1}$ is not bounded. Now, this leads to a contradiction since we assumed that $T$ was open, hence $T^{-1}$ was bounded. It follows that our assumption that $T$ is open cannot be true. It follows that $T$ satisfies the necessary properties.
Is this proof correct?
Best Answer
Consider $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=x^3 -3x^2$. This is bounded (as it's continuous) and surjective. However, the open set $(0,3)$ in the domain maps to $[-4,0)$, which isn't open.
(Aweygan gets the assist.)