A closed subspace $V$ of a normed space $X$ is called complemented if there exist a closed subspace $E$ such that $E \oplus V=X$, equivalently if there exist a continuous operator $T:X\rightarrow V$ such that $T(x)=x \ \ \ \forall x \in V$.
Is $C[0,1]$ complemented in $B[0,1]$ (the space of al bounded functions in $[0,1]$)?
The space of all bounded functions on $[0,1]$ is nothing more than $\ell_\infty(\mathfrak{c})$. The latter space is injective.
Assume $C[0,1]$ is complemented in $B[0,1]$, then it is also injective. Note that any injective Banach space contains a copy of $\ell_\infty$, and therefore non-separable. So $C[0,1]$ must be non-separable too. Contradiction.