I was following the proof of the Open Mapping Theorem in Lang's Real and Functional Analysis and something odd happened. I was able to simplify a lot his proof. Not only that, but I was able to improve the theorem itself. My proof is valid for any normed space, not only Banach spaces. Of course I know this can't be the case, but still, I wasn't able to find where my proof got wrong. If you spot the mistake, please tell me.
Open Mapping Theorem: Let $X,Y$ be Banach spaces and let $f:X \to Y$ be a linear continuous map. If $f$ is surjective, then it is also open.
Proof: For any $s>0$, we will denote $B_s = \{ x\in X:\ \|x\| < s \}$ and $C_s = \{ y \in Y:\ \|y\| < s \}$.
First, note that since $X$ is surjective, we have $$ f(X) = f(\cup_{n=1}^\infty B_n) = Y.$$ Also, note that $s < t \implies B_s \subset B_t$. In particular, this means that $\cup_{n=1}^N B_n \subset B_t$, for any $t > N$. Then for any $r > 0$, there is a $k \in \mathbb{N}$ such that $C_r \subset f(B_{kr})$.
Now let $U \subset X$ be an open set and let $x \in U$ be an arbitrary point. By definition, there is a $t > 0$ such that $x + B_t \subset U$. Then $$f(x+B_t) = f(x) + f(B_t) \subset f(U).$$
Given any $r>0$, we know there is a $k \in \mathbb{K}$ such that $C_r \subset f(B_{kr})$. The it follows that $$f(B_t) = f\left(\frac{t}{kr} B_{kr}\right) = \frac{t}{kr} f(B_{kr}) \supset \frac{t}{kr} C_r = C_\frac{t}{k}.$$
Therefore, $$f(x) + C_\frac{t}{r} \subset f(x) + f(B_t) \subset f(U). $$
This proves $f(x)$ is in the interior of $f(U)$. Since $x$ was arbitrary, we conclude that every point of $f(U)$ is in its interior, so $f(U)$ is open. $\hspace{1cm}\square$
Thank you!
$f(\bigcup_n B_n)=Y$ does not implies there exists $l$ such that $C_r\subset f(B_l)$, the image of a ball is not always a ball.