# Open Mapping Theorem fails when space is not complete

I am working on the open mapping theorem, and I want to show that completeness is a necessary condition for it to work. I have shown that if $$X$$ is Banach, $$Y$$ is a normed space, then there exists a surjective bounded linear operator which is not open. Now, I want to show the other possibility, namely

For $$X$$ a normed space, $$Y$$ Banach, find a surjective bounded linear operator $$T:X\to Y$$ which is not open. I wanted to let $$Y$$ be an infinite dimensional Banach space since I know there exists an unbounded linear function on $$Y$$. What can I take for $$X$$, which map can I use and how can I show this map is not open?

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.