Easy way to extend continous functional There must be a mistake in my reasoning, but I couldn't find it and that's what I ask you to do:
Let $E_0 \subset E$ be a subspace in a normed space $E$ and $f: E_0 \rightarrow \mathbb{R}$ is continous functional  on $E_0$
Let $\{e_0^1, e_0^2, ...\}$ be a basis in $E_0$. We can extend it to the basis of entire $E$ using the Zorn's lemma: let $\{e_1^1, e_1^2,...\}$ be a system that $span \{e_0^1, e_0^2, ..., e_1^1, e_2^2, ...\}=E$
Let's define $f(e_1^i)=0$ for all $i$
So we've extended  continous functional from subspace $E_0$ to the entire space $E$. But it's pretty much easier than how we do it in Hahn–Banach theorem so it must be a mistake somewhere
Maybe the extenstion will not be continious or it's not correct. But I can't understand. Need help to find a mistake

Your extension is in general not continuous, let alone norm preserving, which is what Hahn-Banach gives us.
Consider $f:\mathbb{R}\times\{0\}\to\mathbb{R}$, $f(r,0)=r$ and then the unfortunate choice of basis ${(1,0),(1,\varepsilon)}$. Let $g$ denote your extension. Then $g(0,-n\varepsilon)=n$.
This is clearly not norm preserving. In infinite dimensions this can get so bad that the function is not even bounded.
