Let $\mathcal B$ be Banach. Let $z \in \mathcal B$ be non-zero. Consider $A = \operatorname{span}(\{z\})$. Is it possible to find a closed subspace $C$ such that $$\mathcal B = A \oplus C?$$
If so, how would one proceed to do so? If $\mathcal B$ is also Hilbert, then this would be $C = z^\perp$. Though, it's not clear how this is done in general.
If it is possible, I'm looking for a small hint not a full answer.
For any linear functional $f$ on $X$ and any $z\notin \ker f$, we have $$X=\operatorname{span}(z)\oplus \ker f$$ via the explicit isomorphism $$ x\mapsto \left( \frac{f(x)}{f(z)}z, \ x - \frac{f(x)}{f(z)}z \right) $$ the inverse of which is $(a,b)\mapsto a+b$.
Thus, given $z$, one only needs $f$ such that $f(z)\ne 0$; the Hahn-Banach theorem delivers that (and more).
One says that every one-dimensional subspace is complemented in $X$.