What is wrong with the following trivial proof of the Hahn Banach Theorem?
Hahn Banach Theorem: Let $V$ is a real normed vector space and $U$ a subspace. Then if $\phi : U \rightarrow \mathbb{R}$ is a linear functional bounded by $C>0$, then there exists an extension of $\phi$, $\phi ': V\rightarrow \mathbb{R}$, also linear and bounded by $C$.
"Proof": Since $V/U$ has a basis (by the axiom of choice), it seems we can lift it to get a complimentary subspace $W \subset V$ such that $V= W \oplus U$ (as vector spaces). Then just define $\phi'(w+u)=\phi (u)$.
But this can't be right….
How do You even know that $\phi'$ is continuous? You need to ensure that there exists $C_1$ such that $||u||\leq C_1||w+u||$ for any $w\in W, u\in U$ and to get Hahn-Banach You need $C_1=1$. Your proof works in unitary spaces since You can always take for $W$ an orthogonal complement of $U$.